Chapter 5 Additional Applications of Newton’s...

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Chapter 5 Additional Applications of Newton’s Laws Conceptual Problems

1 • [SSM] Various objects lie on the bed of a truck that is moving along a straight horizontal road. If the truck gradually speeds up, what force acts on the objects to cause them to speed up too? Explain why some of the objects might stay stationary on the floor while others might slip backward on the floor. Determine the Concept Static and kinetic frictional forces are responsible for the accelerations. If the coefficient of static friction between the truck bed and the object is sufficiently large, then the object will not slip on the truck bed. The larger the acceleration of the truck, the larger the coefficient of static friction that is needed to prevent slipping.

2 • Blocks made of the same material but differing in size lie on the bed of a truck that is moving along a straight horizontal road. All of the blocks will slide if the truck’s acceleration is sufficiently great. How does the minimum acceleration at which a small block slips compare with the minimum acceleration at which a much heavier block slips? Determine the Concept The forces acting on an object are the normal force exerted by the floor of the truck, the gravitational force exerted by Earth, and the friction force; also exerted by the floor of the truck. Of these forces, the only one that acts in the direction of the acceleration (chosen to be to the right) is the static friction force. Apply Newton’s second law to the object to determine how the critical acceleration depends on its weight.

x

y

nFr

gFr

sfr

Taking the +x direction to be to the right, apply ΣFx = max to the object:

xmamgFf === sgss μμ ⇒ gax sμ=

Because xa is independent of m and Fg, the critical accelerations are the same. 3 • A block of mass m rests on a plane that is inclined at an angle θ with the horizontal. It follows that the coefficient of static friction between the block and plane is (a) μs ≥ g, (b) μs = tan θ, (c) μs ≤ tan θ, (d) μs ≥ tan θ.

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Determine the Concept The forces acting on the block are the normal force

nFr

exerted by the incline, the weight of the block gF

rexerted by Earth, and the

static friction force sfr

also exerted by the incline. We can use the definition of μs and the conditions for equilibrium to determine the relationship between μs and θ.

sfr

gFr

nFr

θ

x

y

Apply xx maF =∑ to the block: 0sings =− θFf

or, because Fg = mg, 0sins =− θmgf (1)

Apply yy maF =∑ in the y direction:

0cosn =− θmgF (2)

Divide equation (1) by equation (2) to obtain:

n

stanFf

=θ

Substitute for fs (≤ μsFn) and simplify to obtain: s

n

nstan μμθ =≤FF and )(d is correct.

4 • A block of mass m is at rest on a plane that is inclined at an angle of 30º with the horizontal, as shown in Figure 5-56. Which of the following statements about the magnitude of the static frictional force fs is necessarily true? (a) fs > mg. (b) fs > mg cos 30º. (c) fs = mg cos 30º. (d) fs = mg sin 30°. (e) None of these statements is true. Determine the Concept The block is in equilibrium under the influence of ,nF

r,Fg

r and ;sf

r that is,

nFr

+ gFr

+ sfr

= 0 We can apply Newton’s second law in the x direction to determine the relationship between fs and Fg = mg.

sfr

gFr

nFr

x

y

°30

Additional Applications of Newton’s Laws

389

Apply 0=∑ xF to the block to

obtain:

0sins =− θmgf ⇒ θsins mgf =

and )(d is correct.

5 •• On an icy winter day, the coefficient of friction between the tires of a car and a roadway is reduced to one-quarter of its value on a dry day. As a result, the maximum speed vmax dry at which the car can safely negotiate a curve of radius R is reduced. The new value for this speed is (a) vmax dry, (b) 0.71vmax dry, (c) 0.50vmax dry, (d) 0.25vmax dry, (e) reduced by an unknown amount depending on the car’s mass.

Picture the Problem The forces acting on the car as it rounds a curve of radius R at maximum speed are shown on the free-body diagram to the right. The centripetal force is the static friction force exerted by the roadway on the tires. We can apply Newton’s second law to the car to derive an expression for its maximum speed and then compare the speeds under the two friction conditions described.

x

y

nFr

gFr

max s,fr

Apply ∑ = aF

rrm to the car: ∑ ==

RvmfFx

2max

maxs,

and ∑ =−= 0n mgFFy

From the y equation we have: Fn = mg

Express fs,max in terms of Fn in the x equation and solve for vmax to obtain:

gRv smax μ= (1)

When ss 'μμ = :

gRv 'smax' μ= (2)

Dividing equation (2) by equation (1) yields:

s

s

s

s

max

max '''μμ

μμ

==gRgR

vv

Solve for max'v to obtain: max

s

smax

'' vvμμ

=

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Evaluate max'v for s4

1s' μμ = :

maxmaxmax41

max %505.0' vvvv ===

and )(c is correct.

6 •• If it is started properly on the frictionless inside surface of a cone (Figure 5-57), a block is capable of maintaining uniform circular motion. Draw the free-body diagram of the block and identify clearly which force (or forces or force components) is responsible for the centripetal acceleration of the block. Determine the Concept The forces acting on the block are the normal force

nFr

exerted by the surface of the cone and the gravitational force gF

rexerted

by Earth. The horizontal component of

nFr

is responsible for the centripetal force on the block.

nFr

gFr

7 •• Here is an interesting experiment that you can perform at home: take a wooden block and rest it on the floor or some other flat surface. Attach a rubber band to the block and pull gently and steadily on the rubber band in the horizontal direction. Keep your hand moving at constant speed. At some point, the block will start moving, but it will not move smoothly. Instead, it will start moving, stop again, start moving again, stop again, and so on. Explain why the block moves this way. (The start-stop motion is sometimes called ″stick-slip″ motion.) Determine the Concept As the spring is extended, the force exerted by the spring on the block increases. Once that force exceeds the maximum value of the force of static friction, the block will slip. As it does, it will shorten the length of the spring, decreasing the force that the spring exerts. The force of kinetic friction then slows the block to a stop, which starts the cycle over again. 8 • Viewed from an inertial reference frame, an object is seen to be moving in a circle. Which, if any, of the following statements must be true. (a) A nonzero net force acts on the object. (b) The object cannot have a radially outward force acting on it. (c) At least one of the forces acting on the object must point directly toward the center of the circle. (a) True. The velocity of an object moving in a circle is continually changing independently of whether the object’s speed is changing. The change in the velocity vector and the acceleration vector and the net force acting on the object all point toward the center of circle. This center-pointing force is called a centripetal force.


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(b) False. The only condition that must be satisfied in order that the object move along a circular path is that the net force acting on it be radially inward. (c) False. The only condition that must be satisfied in order that the object move along a circular path is that the net force acting on it be radially inward. 9 •• A particle is traveling in a vertical circle at constant speed. One can conclude that the magnitude of its _____ is(are) constant. (a) velocity, (b) acceleration, (c) net force, (d) apparent weight. Determine the Concept A particle traveling in a vertical circle experiences a downward gravitational force plus an additional force that constrains it to move along a circular path. Because the speed of the particle is constant, the magnitude of its velocity is constant. Because the magnitude of its velocity is constant, its acceleration must be constant. Because the magnitude of its acceleration is constant, the magnitude of the net force acting on it must be constant. Therefore,

)(a , )(b , and )(c are correct. 10 •• You place a lightweight piece of iron on a table and hold a small kitchen magnet above the iron at a distance of 1.00 cm. You find that the magnet cannot lift the iron, even though there is obviously a force between the iron and the magnet. Next, again holding the magnet 1.00 cm above the iron, you drop them from arm’s length, releasing them from rest simultaneously. As they fall, the magnet and the piece of iron bang into each other before hitting the floor. (a) Draw free-body diagrams illustrating all of the forces on the magnet and the iron for each demonstration. (b) Explain why the magnet and iron move closer together while they are falling, even though the magnet cannot lift the piece of iron when it is sitting on the table. Determine the Concept We can analyze these demonstrations by drawing force diagrams for each situation. In both diagrams, h denotes ″hand″, g denotes ″gravitational″, m denotes ″magnetic″, and n denotes ″normal.″ (a) Demonstration 1:

Demonstration 2:

(b) Because the magnet doesn’t lift the iron in the first demonstration, the force exerted on the iron must be less than its (the iron’s) weight. This is still true when the two are falling, but the motion of the iron is not restrained by the table, and the

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motion of the magnet is not restrained by the hand. Looking at the second diagram, the net force pulling the magnet down is greater than its weight, implying that its acceleration is greater than g. The opposite is true for the iron: the magnetic force acts upwards, slowing it down, so its acceleration will be less than g. Because of this, the magnet will catch up to the iron piece as they fall. 11 ••• [SSM] The following question is an excellent ″braintwister″ invented by Boris Korsunsky. Two identical blocks are attached by a massless string running over a pulley as shown in Figure 5-58. The rope initially runs over the pulley at the rope’s midpoint, and the surface that block 1 rests on is frictionless. Blocks 1 and 2 are initially at rest when block 2 is released with the string taut and horizontal. Will block 1 hit the pulley before or after block 2 hits the wall? (Assume that the initial distance from block 1 to the pulley is the same as the initial distance from block 2 to the wall.) There is a very simple solution. Picture the Problem The following free-body diagrams show the forces acting on the two objects some time after block 2 is dropped. Note that the tension in the string is the same on both sides of the pulley. The only force pulling block 2 to the left is the horizontal component of the tension. Because this force is smaller than the T and blocks 1 and 2 have the same mass, the acceleration of block 1 to the right will always be greater than the acceleration of block 2 to the left.

x

y

1 2T

T

mg mg

Fn, 1

Because the initial distance from block 1 to the pulley is the same as the initial distance of block 2 to the wall, block 1 will hit the pulley before block 2 hits the wall.

12 •• In class, most professors do the following experiment while discussing the conditions under which air drag can be neglected while analyzing free-fall. First, a flat piece of paper and a small lead weight are dropped next to each other, and clearly the paper’s acceleration is less than that of the lead weight. Then, the paper is crumpled into a small wad and the experiment repeated. Over the distance of a meter or two, it is clear the acceleration of the paper is now very close to that of the lead weight. To your dismay, the professor calls on you to explain why the paper’s acceleration changed so dramatically. Repeat your explanation here!


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Determine the Concept Air drag depends on the frontal area presented. Reducing it by crumpling the paper makes the force of air drag a lot less so that gravity is the most important force. The paper will thus accelerate at approximately g (until speeds are high enough for drag forces to come back into play in spite of the reduced area). 13 •• [SSM] Jim decides to attempt to set a record for terminal speed in skydiving. Using the knowledge he has gained from a physics course, he makes the following plans. He will be dropped from as high an altitude as possible (equipping himself with oxygen), on a warm day and go into a ″knife″ position in which his body is pointed vertically down and his hands are pointed ahead. He will outfit himself with a special sleek helmet and rounded protective clothing. Explain how each of these factors helps Jim attain the record. Determine the Concept Air drag is proportional to the density of air and to the cross-sectional area of the object. On a warm day the air is less dense. The air is also less dense at high altitudes. Pointing his hands results in less area being presented to air drag forces and, hence, reduces them. Rounded and sleek clothing has the same effect as pointing his hands. 14 •• You are sitting in the passenger seat in a car driving around a circular, horizontal, flat racetrack at a high speed. As you sit there, you ″feel″ a ″force″ pushing you toward the outside of the track. What is the true direction of the force acting on you, and where does it come from? (Assume that you do not slide across the seat.) Explain the sensation of an “outward force” on you in terms of the Newtonian perspective. Determine the Concept In your frame of reference (the accelerating reference frame of the car), the direction of the force must point toward the center of the circular path along which you are traveling; that is, in the direction of the centripetal force that keeps you moving in a circle. The friction between you and the seat you are sitting on supplies this force. The reason you seem to be "pushed" to the outside of the curve is that your body’s inertia "wants", in accordance with Newton’s first law (the law of inertia), to keep it moving in a straight line–that is, tangent to the curve. 15 • [SSM] The mass of the moon is only about 1% of that of Earth. Therefore, the force that keeps the moon in its orbit around Earth (a) is much smaller than the gravitational force exerted on the moon by Earth, (b) is much greater than the gravitational force exerted on the moon by Earth, (c) is the gravitational force exerted on the moon by Earth, (d) cannot be answered yet, because we have not yet studied Newton’s law of gravity. Determine the Concept The centripetal force that keeps the moon in its orbit around Earth is provided by the gravitational force Earth exerts on the moon. As described by Newton’s 3rd law, this force is equal in magnitude to the force the

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moon exerts on Earth. )(c is correct. 16 • A block is sliding on a frictionless surface along a loop-the-loop, as in Figure 5-59. The block is moving fast enough so that it never loses contact with the track. Match the points along the track to the appropriate free-body diagrams in the figure. Determine the Concept The only forces acting on the block are its weight and the force the surface exerts on it. Because the loop-the-loop surface is frictionless, the force it exerts on the block must be perpendicular to its surface. At point A the weight is downward and the normal force is to the right. The normal force is the centripetal force. Free-body diagram 3 matches these forces. At point B the weight is downward, the normal force is upward, and the normal force is greater than the weight so that their difference is the centripetal force. Free-body diagram 4 matches these forces. At point C the weight is downward and the normal force is to the left. The normal force is the centripetal force. Free-body diagram 5 matches these forces. At point D both the weight and the normal forces are downward. Their sum is the centripetal force. Free-body diagram 2 matches these forces. 17 •• [SSM] (a) A rock and a feather held at the same height above the ground are simultaneously dropped. During the first few milliseconds following release, the drag force on the rock is smaller than the drag force on the feather, but later on during the fall the opposite is true. Explain. (b) In light of this result, explain how the rock’s acceleration can be so obviously larger than that of the feather. Hint: Draw a free-body diagram of each object. Determine the Concept The drag force


395

acting on the objects is given by ,2

21

d vCAF ρ= where A is the projected surface area, v is the object’s speed, ρ is the density of air, and C is a dimensionless coefficient. We’ll assume that, over the height of the fall, the density of air ρ is constant. The free-body diagrams for a feather and a rock several milliseconds into their fall are shown to the right. The forces acting on both objects are the downward gravitational force and an upward drag force.

F

F

F

F

d, feather

g, feather

d, rock

g, rock

(a) The drag force is proportional to the area presented and some power of the speed. The drag force on the feather is larger because the feather presents a larger area than does the rock. As the rock gains speed, the drag force on it increases. The drag force on the rock eventually exceeds the drag force on the feather because the drag force on the feather cannot exceed the gravitational force on the feather. A short time after being released the feather reaches terminal speed. During the rest of its fall the drag force on it is equal to the gravitational force acting on it. However, the gravitational force on the rock is much greater than the drag force on the feather, so the rock continues to gain speed long after the feather reaches terminal speed. As the rock continues to gain speed, the drag force on it continues to increase. As a result, the drag force on the rock eventually exceeds the drag force on the feather. (b) Terminal speed is much higher for the rock than for the feather. The acceleration of the rock will remain high until its speed approaches its terminal speed. 18 •• Two pucks of masses m1 and m2 are lying on a frictionless table and are connected by a massless spring of force constant k. A horizontal force F1 directed away from m2 is then exerted on m1. What is the magnitude of the resulting acceleration of the center of mass of the two-puck system? (a) F1/m1. (b) F1/(m1 + m2). (c) (F1 + kx)/(m1 + m2), where x is the amount the spring is stretched. (d) (m1 + m2)F1/m1m2.

Determine the Concept The acceleration of the center of mass of a system of particles is described by ,cm

iexti,extnet, aFF

rrrM== ∑ where M is the total mass of the

system.

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Express the acceleration of the center of mass of the two pucks: 21

1extnet,cm mm

FM

Fa

+==

because the spring force is an internal force. )( b is correct.

19 •• The two pucks in Problem 18 lie unconnected on a frictionless table. A horizontal force F1 directed away from m2 is then exerted on m1. How does the magnitude of the resulting acceleration of the center of mass of the two-puck system compare to the magnitude of the acceleration of m1? Explain your reasoning. Determine the Concept The magnitude of the acceleration of the puck whose mass is m1 is related to the net force F1 acting on it through Newton’s second law. Because the pucks are no longer connected, the magnitude of the acceleration of the center of mass is:

1

1eddisconnect CM, m

Fa =

From Problem 18: 21

1connected CM, mm

Fa+

=

Because 21

1

1

1

mmF

mF

+> the magnitude of the acceleration of m1 is greater.

20 •• If only external forces can cause the center of mass of a system of particles to accelerate, how can a car on level ground ever accelerate? We normally think of the car’s engine as supplying the force needed to accelerate the car, but is this true? Where does the external force that accelerates the car come from? Determine the Concept There is only one force that can cause the car to move forward−the friction of the road! The car’s engine causes the tires to rotate, but if the road were frictionless (as is closely approximated by icy conditions) the wheels would simply spin without the car moving anywhere. Because of friction, the car’s tire pushes backwards against the road−from Newton’s third law, the frictional force acting on the tire must then push it forward. This may seem odd, as we tend to think of friction as being a retarding force only, but it is true. 21 •• When we push on the brake pedal to slow down a car, a brake pad is pressed against the rotor so that the friction of the pad slows the wheel’s rotation. However, the friction of the pad against the rotor cannot be the force that slows the car down, because it is an internal force (both the rotor and the wheel are parts of the car, so any forces between them are purely internal to the system). What is


397

the external force that slows down the car? Give a detailed explanation of how this force operates. Determine the Concept The friction of the tire against the road causes the car to slow down. This is rather subtle, as the tire is in contact with the ground without slipping at all times, and so as you push on the brakes harder, the force of static friction of the road against the tires must increase. 22 •• Give an example of each of the following. (a) A three-dimensional object that has no matter at its center of mass. (b) A solid object whose center of mass is outside of it. (c) A solid sphere whose center of mass does not lie at its geometrical center. (d) A long wooden stick whose center of mass does not lie at its middle. (a) A solid spherical shell, or donut, or tire. (b) A solid hemispherical shell. (c) Any sphere with one side a different density than the other, or a density variation that isn’t radially symmetric. (d) Any stick with a non-uniform and non-symmetric density variation. A baseball bat is a good example of such a stick. 23 •• [SSM] When you are standing upright, your center of mass is located within the volume of your body. However, as you bend over (say to pick up a package), its location changes. Approximately where is it when you are bent over at right angles and what change in your body caused the center of mass location to change? Explain. Determine the Concept Relative to the ground, your center of mass moves downward. This is because some of your mass (hips) moved backward, some of your mass (your head and shoulders) moved forward, and the top half of your body moved downward. 24 •• Early on their three-day (one-way) trip to the moon, the Apollo team (late 1960s to early 1970s) would explosively separate the lunar ship from the third-stage booster (that provided the final ″boost″) while still fairly close to Earth. During the explosion, how did the velocity of each of the two pieces of the system change? How did the velocity of the center of mass of the system change? Determine the Concept The spacecraft speed increased toward the moon. The speed of the third-stage booster decreased, but the booster continued to move away from Earth and toward the moon. Right after the explosion, the center of

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mass velocity was the same as before. After a while, however, the backward pull of gravity of Earth will cause it to decrease because the speeds of both the lunar ship and the booster decrease. 25 •• You throw a boomerang and for a while it ″flies″ horizontally in a straight line at a constant speed, while spinning rapidly. Draw a series of pictures, as viewed vertically down from overhead, of the boomerang in different rotational positions as it moves parallel to the surface of Earth. On each picture, indicate the location of the boomerang’s center of mass and connect the dots to trace the trajectory of its center of mass. What is the center of mass’s acceleration during this part of the flight? Determine the Concept The diagram shows a spinning boomerang with its center of mass at the location of the circle. As viewed from above, the center of mass moves in a straight line as the boomerang spins about it. The acceleration of the center of mass is zero.

+ ++ ++

Estimation and Approximation 26 •• To determine the aerodynamic drag on a car, automotive engineers often use the ″coast-down″ method. The car is driven on a long, flat road at some convenient speed (60 mi/h is typical), shifted into neutral, and allowed to coast to a stop. The time that it takes for the speed to drop by successive 5-mi/h intervals is measured and used to compute the net force slowing the car down. (a) One day, a group measured that a Toyota Tercel with a mass of 1020 kg coasted down from 60.0 mi/h to 55.0 mi/h in 3.92 s. Estimate the average net force slowing the car down in this speed range. (b) If the coefficient of rolling friction for this car is known to be 0.020, what is the force of rolling friction that is acting to slow it down? Assuming that the only two forces acting on the car are rolling friction and aerodynamic drag, what is the average drag force acting on the car? (c) The drag force has the form 21

2C Avρ , where A is the cross-sectional area of the car facing into the air, v is the car’s speed, ρ is the density of air, and C is a dimensionless constant of order 1. If the cross-sectional area of the car is 1.91 m2, determine C from the data given. (The density of air is 1.21 kg/m3; use 57.5 mi/h for the speed of the car in this computation.) Picture the Problem The forces acting on the Tercel as it slows from 60 to 55 mi/h are a rolling-friction force exerted by the roadway, an air-drag force exerted by the air, the normal force exerted by the roadway, and the gravitational force exerted by Earth. The car is moving in the positive x direction. We can use Newton’s second law to calculate the average force from the rate at which the


399

car’s speed decreases and the rolling force from its definition. The drag force can be inferred from the average- and rolling-friction forces and the drag coefficient from the defining equation for the drag force.

x

y

nFr

gFr

rollingfr

dFr

(a) Apply∑ = xx maF to the car to relate the average force acting on it to its average acceleration:

tvmmaFΔΔ

== avav

Substitute numerical values and evaluate avF :

( ) kN58.0N581s3.92

kmm1000

s3600h1

mikm1.609

hmi5

kg1020av ==×××

=F

(b) The rolling-friction force is the product of the coefficient of rolling friction and the normal force:

mgFf rollingnrollingrolling μμ ==

Substitute numerical values and evaluate rollingf :

( )( )( )kN20.0

m/s9.81kg1020020.0 2rolling

=

=f

Assuming that only two forces are acting on the car opposite to the direction of its motion gives:

rollingdav fFF += ⇒ rollingavd fFF −=

Substitute numerical values and evaluate Fd:

kN38.0N200N815d =−=F

(c) Using the definition of the drag force and its calculated value from (b) and the average speed of the car during this 5 mph interval, solve for C:

221

d AvCF ρ= ⇒ 2

d2

Av

FC

ρ=

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Substitute numerical values and evaluate C:

( )

( )( )50.0

kmm10

s3600h1

mikm1.609

hmi57.5m1.91kg/m1.21

N381223

23

=

⎟⎟⎠

⎞⎜⎜⎝

⎛×××

=C

27 •• [SSM] Using dimensional analysis, determine the units and dimensions of the constant b in the retarding force bvn if (a) n = 1 and (b) n = 2. (c) Newton showed that the air resistance of a falling object with a circular cross section should be approximately 2 21

2 r vρπ , where ρ = 1.20 kg/m3, the density of air. Show that this is consistent with your dimensional analysis for Part (b). (d) Find the terminal speed for a 56.0-kg skydiver; approximate his cross-sectional area as a disk of radius 0.30 m. The density of air near the surface of Earth is 1.20 kg/m3. (e) The density of the atmosphere decreases with height above the surface of Earth; at a height of 8.0 km, the density is only 0.514 kg/m3. What is the terminal velocity at this height? Picture the Problem We can use the dimensions of force and speed to determine the dimensions of the constant b and the dimensions of ρ, r, and v to show that, for n = 2, Newton’s expression is consistent dimensionally with our result from part (b). In Parts (d) and (e), we can apply Newton’s second law under terminal speed conditions to find the terminal speed of the sky diver near the surface of Earth and at a height of 8 km. Assume that g = 9.81 m/s2 remains constant. (Note: At 8 km, g = 9.78 m/s2. However, it will not affect the result in Part (e).) (a) Solve the drag force equation for b with n = 1:

vFb d=

Substitute the dimensions of Fd and v and simplify to obtain: [ ]

TM

TL

TML

2==b

and the units of b are kg/s

(b) Solve the drag force equation for b with n = 2:

2d

vFb =

Substitute the dimensions of Fd and v and simplify to obtain: [ ]

LM

TLTML

2

2=

⎟⎠⎞

⎜⎝⎛

=b

and the units of b are kg/m


401

(c) Express the dimensions of Newton’s expression:

[ ] [ ] ( )

2

22

322

21

d

TML

TLL

LM

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛== vrF ρπ

From Part (b) we have: [ ] [ ] 2

22

d TML

TL

LM

=⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛== bvF

(d) Letting the downward direction be the +y direction, apply ∑ = yy maF to the sky diver:

02t

221 =− vrmg ρπ ⇒ 2t

2r

mgvρπ

=

Substitute numerical values and evaluate vt:

( )( )( )( )

m/s57m0.30kg/m1.2

m/s9.81kg56223

2

t ==π

v

(e) Evaluate vt at a height of 8 km: ( )( )

( )( )m/s87

m0.30kg/m514.0m/s9.81kg562

23

2

t

=

=π

v

28 •• Estimate the terminal velocity of an average sized raindrop and a golf-ball- sized hailstone. (Hint: See Problems 26 and 27.) Picture the Problem From Newton’s second law, the equation describing the motion of falling raindrops and large hailstones is mg – Fd = ma where

22221

d bvvrF == ρπ is the drag force. Under terminal speed conditions (a = 0), the drag force is equal to the weight of the falling object. Take the radius of a raindrop to be 0.50 mm and the radius of a golf-ball sized hailstone to be 2.0 cm. Express the relationship between vt and the weight of a falling object under terminal speed:

mgbvt =2 ⇒

bmgv =t (1)

Using 221 rb πρ= , evaluate br: ( )( )

kg/m1071.4

m1050.0kg/m2.17

23321

r

−

−

×=

×= πb

Evaluating bh yields: ( )( )

kg/m1054.7

m100.2kg/m2.14

22321

h

−

−

×=

×= πb

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Express the mass of a sphere in terms of its volume and density: 3

4 3ρπρ rVm ==

Using ρr = 1.0 × 103 kg/m3, evaluate mr:

( ) ( )

kg1024.53

kg/m100.1m1050.04

7

3333

r

−

−

×=

××=

πm

Using ρh = 920 kg/m3, evaluate mh: ( ) ( )

kg1008.33

kg/m920m100.24

2

332

h

−

−

×=

×=

πm

Substitute numerical values in equation (1) and evaluate vt,r:

( )( )

m/s3.3

kg/m104.71m/s9.81kg105.24

7

27

rt,

=

××

= −

−

v

Substitute numerical values in equation (1) and evaluate vt,h:

( )( )

m/s20

kg/m1054.7m/s9.81kg1008.3

4

22

ht,

=

××

= −

−

v

29 •• Estimate the minimum coefficient of static friction needed between a car’s tires and the pavement in order to complete a left turn at a city street intersection at the posted straight-ahead speed limit of 25 mph and on narrow inner-city streets. Comment on the wisdom of attempting such a turn at that speed. Picture the Problem In order to perform this estimate, we need to determine a rough radius of curvature for the car’s turn in a normal city intersection. Assuming the car goes from right-hand lane to right-hand lane, and assuming fairly normal dimensions of 40 feet for the width of the street, the center of the car’s path travels along a circle of, say, 30 feet in radius. The net centripetal force is provided by the force of static friction and the acceleration of the car is equal to this net force divided by the mass of the car. Finally, we solve for the coefficient of static friction.


403

A diagram showing the forces acting on the car as it rounds the curve is shown to the right.

x

y

nFr

gFr

max s,fr

Apply xx maF =∑ to the car’s tires: r

vmmaf x

2

max s, ==

or, because nsmax s, Ff μ=

rvmF

2

ns =μ ⇒ n

2

s rFmv

=μ (1)

Apply yy maF =∑ to the car’s tires:

0gn =− FF

or, because Fg = mg, 0n =−mgF ⇒ mgF =n

Substituting for Fn in equation (1) yields: rg

vrmgmv 22

s ==μ

Substitute numerical values and evaluate μs:

( )

4.1

m/s 81.9ftm 0.3048ft 30

mim 1609

s 3600h 1

hmi 25

2

2

s

=

⎟⎠⎞

⎜⎝⎛ ×

⎟⎠⎞

⎜⎝⎛ ××

=μ

This is probably not such a good idea. Tires on asphalt or concrete have a maximum coefficient of static friction of about 1. 30 •• Estimate the widest stance you can take when standing on a dry, icy surface. That is, how wide can you safely place your feet and not slip into an undesired ″split?″ Let the coefficient of static friction of rubber on ice be roughly 0.25.

Picture the Problem We need to estimate the forces active at the place of each foot. Assuming a symmetrical stance, with the defining angle being the angle between each leg and the ground,θ, we can then draw a force diagram and apply Newton’s second law to your foot. The free-body diagram shows the normal

Chapter 5

404

force, exerted by the icy surface, the maximum static friction force, also exerted by the icy surface, and the force due to gravity exerted on your foot.

A free-body diagram showing the forces acting on one foot is shown to the right.

x

y

nFr

max s,frθ

gFr

Apply aFrr

m=∑ to one of your feet:

0cosgmaxs, =−=∑ θFfFx (1)

and

0singn =−=∑ θFFFy (2)

Substituting fs, max = μsFn in equation (1) gives:

0cosgns =− θμ FF

or

θμ cosgns FF = (3)

Solving equation (2) for Fn yields:

θsingn FF = (4)

Divide equation (4) by equation (3) to obtain: s

1tanμ

θ = ⇒ ⎟⎟⎠

⎞⎜⎜⎝

⎛= −

s

1 1tanμ

θ

Substitute the numerical value of sμ

and evaluate θ: °=⎟

⎠⎞

⎜⎝⎛= − 76

25.01tan 1θ

This angle corresponds to an angle between your legs of about 28°.

Friction 31 • [SSM] A block of mass m slides at constant speed down a plane inclined at an angle of θ with the horizontal. It follows that (a) μk = mg sin θ, (b) μk = tan θ, (c) μk = 1 – cos θ, (d) μk = cos θ – sin θ. Picture the Problem The block is in equilibrium under the influence of nF

r, ,mg

r

and ;kfr

that is, nFr

+ gr

m + kfr

= 0. We can apply Newton’s second law to determine the relationship between fk, θ, and mg.


405

A pictorial representation showing the forces acting on the sliding block is shown to the right.

gFr

nFr

θ

x

y

kfr

Using its definition, express the coefficient of kinetic friction:

n

kk F

f=μ (1)

Apply ∑ = xx maF to the block: xmamgf =− θsink or, because ax = 0,

θsink mgf =

Apply ∑ = yy maF to the block: ymamgF =− θcosn

or, because ay = 0, θcosn mgF =

Substitute for fk and Fn in equation (1) and simplify to obtain:

θθθμ tan

cossin

k ==mgmg

and )(b is correct. 32 • A block of wood is pulled at constant velocity by a horizontal string across a horizontal surface with a force of 20 N. The coefficient of kinetic friction between the surfaces is 0.3. The force of friction is (a) impossible to determine without knowing the mass of the block, (b) impossible to determine without knowing the speed of the block, (c) 0.30 N, (d) 6.0 N, or (e) 20 N. Picture the Problem The block is in equilibrium under the influence of nF

r,

,gFr

,appFr

and ;kfr

that is

nFr

+ gFr

+ appFr

+ kfr

= 0

We can apply Newton’s second law to determine fk.

x

y

nFr

gFr

appFr

kfr

Chapter 5

406

Apply ∑ = xx maF to the block: xmafF =− kapp

or, because ax = 0, N20appk == Ff and )(e is correct.

33 • [SSM] A block weighing 20-N rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block are μs = 0.80 and μk = 0.60. A horizontal string is then attached to the block and a constant tension T is maintained in the string. What is the subsequent force of friction acting on the block if (a) T = 15 N or (b) T = 20 N?

Picture the Problem Whether the friction force is that due to static friction or kinetic friction depends on whether the applied tension is greater than the maximum static friction force. We can apply the definition of the maximum static friction to decide whether fs,max or T is greater.

x

y

nFr

gFrkf

r

Tr

Noting that gn FF = , calculate the

maximum static friction force:

( )( )N16

N 2080.0gsnsmaxs,

=

=== FFf μμ

(a) Because fs,max > T: N 15s === Tff

(b) Because T > fs,max:

( )( ) N 12N 2060.0

gknkk

==

=== FFff μμ

34 • A block of mass m is pulled at a constant velocity across a horizontal surface by a string as shown in Figure 5-60. The magnitude of the frictional force is (a) μk mg, (b) T cos θ, (c) μk (T – mg), (d) μk T sin θ, or (e) μk (mg – T sin θ). Picture the Problem The block is in equilibrium under the influence of the forces

,Tr

,kfr

,nFr

and gFr

; that is Tr

+ kfr

+ gFr

+ nFr

= 0. We can apply Newton’s second law to determine the relationship between T and fk.


407

A free-body diagram showing the forces acting on the block is shown to the right.

x

y

nFr

gFr kf

r

Tr

θ

Apply ∑ = xx maF to the block: xmafT =+− kcosθ

Because ax = 0: θcosk Tf = and )(b is correct. 35 • [SSM] A 100-kg crate rests on a thick-pile carpet. A weary worker then pushes on the crate with a horizontal force of 500 N. The coefficients of static and kinetic friction between the crate and the carpet are 0.600 and 0.400, respectively. Find the subsequent frictional force exerted by the carpet on the crate. Picture the Problem Whether the friction force is that due to static friction or kinetic friction depends on whether the applied tension is greater than the maximum static friction force. If it is, then the box moves and the friction force is the force of kinetic friction. If it is less, the box does not move.

x

y

nFr

gFr

appFr

fr

The maximum static friction force is given by:

nsmaxs, Ff μ=

or, because Fn = Fg = mg, mgf smaxs, μ=

Substitute numerical values and evaluate fs,max:

( )( )( )N589

m/s81.9kg100600.0 2maxs,

=

=f

Because appmax s, Ff > , the box does

not move and : N500app == sfF

36 • A box weighing 600 N is pushed along a horizontal floor at constant velocity with a force of 250 N parallel to the floor. What is the coefficient of kinetic friction between the box and the floor?

Chapter 5

408

Picture the Problem Because the box is moving with constant velocity, its acceleration is zero and it is in equilibrium under the influence of ,appF

r,nF

r ,gFr

and kfr

;

that is, appFr

+ nFr

+ gFr

+ kfr

= 0. We

can apply Newton’s second law to determine the relationship between fk and mg.

x

y

nFr

gFr

appFr

kfr

The definition of μk is:

n

kk F

f=μ (1)

Apply ∑ = yy maF to the box: ymaFF =− gn

or, because ay = 0, 0gn =− FF ⇒ N 600gn === mgFF

Apply ∑ = xx maF to the box: xmafF =− kapp

or, because ax = 0, N250kapp == fF

Substitute numerical values in equation (1) and evaluate μk:

417.0N600N250

k ==μ

37 • [SSM] The coefficient of static friction between the tires of a car and a horizontal road is 0.60. Neglecting air resistance and rolling friction, (a) what is the magnitude of the maximum acceleration of the car when it is braked? (b) What is the shortest distance in which the car can stop if it is initially traveling at 30 m/s? Picture the Problem Assume that the car is traveling to the right and let the positive x direction also be to the right. We can use Newton’s second law of motion and the definition of μs to determine the maximum acceleration of the car. Once we know the car’s maximum acceleration, we can use a constant-acceleration equation to determine the least stopping distance.


409

(a) A diagram showing the forces acting on the car is shown to the right.

nFr

gFr

kfr

y

x

Apply∑ = xx maF to the car: xmaFf =−=− nsmaxs, μ (1)

Apply∑ = yy maF to the car and

solve for Fn: ymaFF =− gn

Fn − w = may = 0 or, because ay = 0 and Fg = mg,

mgF =n (2)

Substitute for Fn in equation (1) to obtain:

xmamgf =−=− smaxs, μ

Solving for max,xa yields: = smax, gax μ

Substitute numerical values and evaluate max,xa : 2

22max,

m/s5.9

m/s5.89=)m/s 1(0.60)(9.8 =

=

xa

(b) Using a constant-acceleration equation, relate the stopping distance of the car to its initial speed and its acceleration:

xavv xxx Δ220

2 += or, because vx = 0,

xav xx Δ20 20 += ⇒

x

x

avx

2Δ

20−

=

Using ax = −5.89 m/s2 because the acceleration of the car is to the left, substitute numerical values and evaluate Δx:

( )( ) m76

m/s89.52m/s30Δ 2

2

=−−

=x

38 • The force that accelerates a car along a flat road is the frictional force exerted by the road on the car’s tires. (a) Explain why the acceleration can be greater when the wheels do not slip. (b) If a car is to accelerate from 0 to 90 km/h in 12 s, what is the minimum coefficient of friction needed between the road and tires? Assume that the drive wheels support exactly half the weight of the car.

Chapter 5

410

Picture the Problem We can use the definition of acceleration and apply Newton’s second law to the horizontal and vertical components of the forces to determine the minimum coefficient of friction between the road and the tires. (a) The free-body diagram shows the forces acting on the tires on the drive wheels, the tires we’re assuming support half the weight of the car.

x

y

nFr

gFr

sfr

Because ks μμ > , f will be greater if the wheels do not slip. (b) Apply∑ = xx maF to the car: xmaFf == nss μ (1)

Apply∑ = yy maF to the car and solve for Fn:

ymaFF =− gn or, because ay = 0 and ,2

1g mgF =

mgF 21

n =

Substituting for Fn in equation (1) yields:

xmamg =s21 μ ⇒

gax2

s =μ

Substitute for ax to obtain:

tgvΔΔ2

s =μ


( )

0.42

s12sm 81.9

s 3600h 1

kmm 10

hkm902

2

3

s

=

⎟⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛××

=μ

39 •• A 5.00-kg block is held at rest against a vertical wall by a horizontal force of 100 N. (a) What is the frictional force exerted by the wall on the block? (b) What is the minimum horizontal force needed to prevent the block from falling if the static coefficient of friction between the wall and the block is 0.400?


411

Picture the Problem The block is in equilibrium under the influence of the forces shown on the force diagram. We can use Newton’s second law and the definition of μs to solve for fs and Fn.

x

y

nFr

gFr

100 N

sfr

(a) Apply∑ = yy maF to the

block: ymamgf =−s

or, because ay = 0, 0s =−mgf ⇒ mgf =s

Substitute numerical values and evaluate fs :

( )( ) N1.49m/s81.9kg00.5 2s ==f

(b) Use the definition of μs to express Fn: s

maxs,n μ

fF =

Substitute numerical values and evaluate Fn:

N1230.400

N1.49n ==F

40 •• A tired and overloaded student is attempting to hold a large physics textbook wedged under his arm, as shown in Figure 5-61. The textbook has a mass of 3.2 kg, while the coefficient of static friction of the textbook against the student’s underarm is 0.320 and the coefficient of static friction of the book against the student’s shirt is 0.160. (a) What is the minimum horizontal force that the student must apply to the textbook to prevent it from falling? (b) If the student can only exert a force of 61 N, what is the acceleration of the textbook as it slides from under his arm? The coefficient of kinetic friction of arm against textbook is 0.200, while that of shirt against textbook is 0.090. Picture the Problem We can apply Newton’s second law to relate the minimum force required to hold the book in place to its mass and to the coefficients of static friction. In Part (b), we can proceed similarly to relate the acceleration of the book to the coefficients of kinetic friction.

Chapter 5

412

(a) The force diagram shows the forces acting on the book. The normal force is the net force the student exerts in squeezing the book. Let the horizontal direction be the +x direction and upward the +y direction. Note that the normal force is the same on either side of the book because it is not accelerating in the horizontal direction. The book could be accelerating downward.

min,1Fr

min,2Fr

min,2Fmin,1F

gmr

μ μ12

Tipl

er/M

osca

x

y

Apply ∑ = aFrr

m to the book:

0and

0

min,2s,2min,11,s

min,1min,2

=−+=

=−=

∑

∑

mgFFF

FFF

y

x

μμ

Noting that ,min,2min,1 FF = solve the y equation for Fmin: s,21 s,

min μμ +=

mgF

Substitute numerical values and evaluate Fmin:

( )( ) N650.1600.320

m/s9.81kg3.2 2

min =+

=F

(b) Apply ∑ = yy maF with the book accelerating downward, to obtain:

mamgFFF ky =−+=∑ k,21, μμ

Solving for ay yields: gF

ma kk

y −+

= 2,1, μμ

Substitute numerical values and evaluate ay: ( )

downward ,m/s3.4

m/s9.81N16kg3.20.0900.200

2

2

=

−⎟⎟⎠

⎞⎜⎜⎝

⎛ +=ya

41 •• You are racing in a rally on a snowy day when the temperature is near the freezing point. The coefficient of static friction between a car’s tires and an icy road is 0.080. Your crew boss is concerned about some of the hills on the course and wants you to think about switching to studded tires. To address the issue, he wants to compare the actual hill angles on the course to see which of them your car can negotiate. (a) What is the angle of the steepest incline that a vehicle with four-wheel drive can climb at constant speed? (b) Given that the hills are icy, what is the steepest possible hill angle for the same four-wheel drive car to descend at constant speed? Picture the Problem We can use the definition of the coefficient of static friction


413

and Newton’s second law to relate the angle of the incline to the forces acting on the car.

(a) The free-body diagram shows the forces acting on the car when it is either moving up the hill or down the hill without acceleration. The friction force that the ground exerts on the tires is the force fs shown acting up the incline.

sfr

gFr

nFr

θ

x

y

Apply aF

rrm=∑ to the car: 0sings =−=∑ θFfFx (1)

and 0cosgn =−=∑ θFFFy (2)

Because Fg = mg, equations (1) and (2) become:

0sins =− θmgf (3) and


Solving equation (3) for fs and equation (4) for Fn yields:

θsins mgf = and

θcosn mgF =

Use the definition of μs to relate fs and Fn:

θθθμ tan

cossin

n

ss ===

mgmg

Ff

Solving forθ yields: ( ) ( ) °=== −− 6.4080.0tantan 1s

1 μθ

(b) Proceed exactly as in (a) to obtain:

( ) °== − 6.4080.0tan 1θ

42 •• A 50-kg box that is resting on a level floor must be moved. The coefficient of static friction between the box and the floor is 0.60. One way to move the box is to push down on the box at an angle θ below the horizontal. Another method is to pull up on the box at an angle θ above the horizontal. (a) Explain why one method requires less force than the other. (b) Calculate the minimum force needed to move the box by each method if θ = 30º and compare the answer with the results when θ = 0°. Picture the Problem The free-body

Chapter 5

414

diagrams for the two methods are shown to the right. Method 1 results in the box being pushed into the floor, increasing the normal force and the static friction force. Method 2 partially lifts the box,, reducing the normal force and the static friction force. We can apply Newton’s second law to obtain expressions that relate the maximum static friction force to the applied force .F

r

x

y

nFr

sfr

Fr

θ

Fr

g

x

y

nFr

sfr

Fr

θ

Fr

g

Method 1 Method 2

(a) Method 2 is preferable because it reduces Fn and, therefore, fs. (b) Apply∑ = xx maF to the box: 0coscos ns =−=− FFfF x μθθ

Method 1: Apply ∑ = yy maF to the

block and solve for Fn:

0sinn =−− θFmgF and

θsinn FmgF +=

Relate maxs,f to Fn: ( )θμμ sinsnsmaxs, FmgFf +== (1) Method 2: Apply ∑ = yy maF to the

forces in the y direction and solve for Fn:

0sinn =+− θFmgF and

θsinn FmgF −=

Relate maxs,f to Fn: ( )θμμ sinssmaxs, FmgFf n −== (2)

Express the condition that must be satisfied to move the box by either method:

θcosmaxs, Ff = (3)

Method 1: Substitute (1) in (3) and solve for F:

θμθμ

sincos s

s1 −=

mgF (4)

Method 2: Substitute (2) in (3) and solve for F: θμθ

μsincos s

s2 +=

mgF (5)


415

Substitute numerical values and evaluate equations (4) and (5) with θ = 30°:

( ) ( )( )( )( )

kN 0.52

30sin60.030cosm/s 81.9kg 5060.030

2

1

=

°−°=°F

and

( ) ( )( )( )( )

kN 0.25

30sin60.030cosm/s 81.9kg 5060.030

2

2

=

°+°=°F

Evaluate equations (4) and (5) with θ = 0°: ( ) ( )( )( )

( )kN 0.29

0sin60.00cosm/s 81.9kg 5060.00

2

1

=

°−°=°F

and

( ) ( )( )( )( )

kN 0.29

0sin60.00cosm/s 81.9kg 5060.00

2

2

=

°+°=°F

43 •• [SSM] A block of mass m1 = 250 g is at rest on a plane that makes an angle of θ = 30° with the horizontal. The coefficient of kinetic friction between the block and the plane is 0.100. The block is attached to a second block of mass m2 = 200 g that hangs freely by a string that passes over a frictionless, massless pulley (Figure 5-62). When the second block has fallen 30.0 cm, what will be its speed? Picture the Problem Choose a coordinate system in which the +x direction is up the incline for the block whose mass is m1 and downward for the block whose mass is m2. We can find the speed of the system when it has moved a given distance by using a constant-acceleration equation. We’ll assume that the string is massless and that it does not stretch. Under the influence of the forces shown in the free-body diagrams, the blocks will have a common acceleration a. The application of Newton’s second law to each block, followed by the elimination of the tension T and the use of the definition of fk, will allow us to determine the acceleration of the system.

x

y

°30

°30

x

1m 2mT

T n,1F

g,1F g,2F kf

Chapter 5

416

Using a constant-acceleration equation, relate the speed of the system to its acceleration and displacement:

Δ220

2 xavv xxx += and, because v0x = 0,

Δ22 xav xx = ⇒ xav xx Δ2= (1)

Apply aFrr

m=∑ to the block whose

mass is m1:

∑ =°−−= xx amFfTF 1g,1k 30sin (2)

and ∑ =°−= 030cosg,1n,1 FFFy (3)

Because Fg,1 = m1g, equations (2) and (3) can be written as:

xamgmfT 11k 30sin =°−− (4) and

°= 30cos1n,1 gmF (5)

Using fk = μkFn,1, substitute equation (5) in equation (4) to obtain:

xamgmgmT

1

11k 30sin30cos=

°−°− μ (6)

Applying∑ = xx maF to the block

whose mass is m2 gives:

xamTF 2g,2 =−

or, because Fg,2 = m2g, xamTgm 22 =− (7)

Add equations (6) and (7) to eliminate T and then solve for ax to obtain:

( )21

11k2 30sin30cosmm

gmmmax +°−°−

=μ

Substituting for ax in equation (1) and simplifying yields:

( )[ ]21

k12 Δ30sin30cos2mm

xgmmvx +°+°−

=μ


417

Substitute numerical values and evaluate vx:

( ) ( )( )[ ]( )( )

cm/s 84

kg 200.0kg 250.0m 300.0m/s 81.930sin30cos100.0kg 250.0kg 200.02 2

=

+°+°−

=xv

44 •• In Figure 5-62 m1 = 4.0 kg and the coefficient of static friction between the block and the incline is 0.40. (a) Find the range of possible values for m2 for which the system will be in static equilibrium. (b) Find the frictional force on the 4.0-kg block if m2 = 1.0 kg? Picture the Problem Choose a coordinate system in which the +x direction is up the incline for the block whose mass is m1 and downward for the block whose mass is m2. We’ll assume that the string is massless and that it does not stretch. Under the influence of the forces shown in the free-body diagrams, the blocks are in static equilibrium. While fs can be either up or down the incline, the free-body diagram shows the situation in which motion is impending up the incline. The application of Newton’s second law to each block, followed by the elimination of the tension T and the use of the definition of fs, will allow us to determine the range of values for m2.

sfr

x

y

n,1Fr

g,1Fr °30

Tr

g,2Fr

x

Tr

m

m

1

2

(a) Noting that ,1g,1 gmF = apply

aFrr

m=∑ to the block whose mass

is m1:

030sin1maxs, =°−±=∑ gmfTFx (1)

and 030cos1n,1 =°−=∑ gmFFy (2)

Using nsmax s, Ff μ= , substitute

equation (2) in equation (1) to obtain:

030sin30cos 11 =°−°± gmgmT sμ (3)

Noting that ,2g,2 gmF = apply

∑ = xx maF to the block whose mass

is m2:

02 =−Tgm (4)

Chapter 5

418

Add equations (3) and (4) to eliminate T and solve for m2:

( )°+°±= 30sin30coss12 μmm (5)

Substitute numerical values to obtain:

( ) ( )[ ]°+°±= 30sin30cos40.0kg0.42m

Denoting the value of m2 with a plus sign as m2,+ and the value of m2 with the minus sign as m2,- determine the range of values of m2 for which the system is in static equilibrium:

kg61.0mandkg4.3 2,-,2 ==+m

and kg4.3kg61.0 2 ≤≤ m

(b) With m2 = 1 kg, the impending motion is down the incline and the static friction force is up the incline. Apply∑ = xx maF to the block

whose mass is m1:

030sin1s =°−+ gmfT (6)

Apply∑ = xx maF to the block

whose mass is m2:

m2g – T = 0 (7)

Add equations (6) and (7) and solve for fs to obtain:

( )gmmf 21s sin30 −°=

Substitute numerical values and evaluate fs:

( )[ ]( )N8.9

m/s81.9kg0.130sinkg0.4 2s

=

−°=f

45 •• In Figure 5-62, m1 = 4.0 kg, m2 = 5.0 kg, and the coefficient of kinetic friction between the inclined plane and the 4.0-kg block is μk = 0.24. Find the magnitude of the acceleration of the masses and the tension in the cord. Picture the Problem Choose a coordinate system in which the +x direction is up the incline for the block whose mass is m1 and downward for the block whose mass is m2. We’ll assume that the string is massless and that it does not stretch. Under the influence of the forces shown in the free-body diagrams, the blocks will have a common acceleration a. The application of Newton’s second law to each block, followed by the elimination of the tension T and the use of the definition of fk, will allow us to determine the acceleration of the system. Finally, we can substitute for the tension in either of the motion equations to determine the acceleration of the masses.


419

sfr

x

y

n,1Fr

g,1Fr °30

Tr

g,2Fr

x

Tr

m

m

1

2

Noting that ,1g,1 gmF = apply

aFrr

m=∑ to the block whose mass

is m1:

xx amgmfTF 11k 30sin =°−−=∑ (1)

and 030cos1n,1 =°−=∑ gmFFy (2)

Using nkk Ff μ= , substitute equation (2) in equation (1) to obtain:

xamgmgmT

11

1k

30sin30cos

=°−°− μ

(3)

Apply∑ = xx maF to the block whose

mass is m2:

xamTgm 22 =− (4)

Add equations (3) and (4) to eliminate T and solve for ax to obtain:

( )21

11k2 30sin30cosmm

gmmmax +°−°−

=μ

Substituting numerical values and evaluating ax yields:

( )( ) ( )[ ] ( ) 22

m/s4.2kg 0.5kg 0.4

m/s 81.930sinkg 0.430coskg 0.424.0kg 0.5=

+°−°−

=xa

Solving equation (3) for T yields:

[ ] gmamT x 1k1 30sin30cos °+°+= μ

Substitute numerical values and evaluate T:

( )( ) ( )[ ]( )( ) N 37m/s 81.9kg 0.430sin30cos24.0m/s 36.2kg 0.4 22 =°+°+=T 46 •• A 12-kg turtle rests on the bed of a zookeeper’s truck, which is traveling down a country road at 55 mi/h. The zookeeper spots a deer in the road, and slows to a stop in 12 s. Assuming constant acceleration, what is the minimum coefficient of static friction between the turtle and the truck bed surface that is needed to prevent the turtle from sliding?

Chapter 5

420

Picture the Problem We can determine the acceleration necessary for the truck and turtle by considering the displacement of both during the given time interval. The static friction force must provide the necessary acceleration for the turtle. The turtle, if it is not to slip, must have this acceleration which is produced by the static friction force acting on it

x

y

nFr

gFr

frs

The required coefficient of static friction is given by:

n

ss F

f=μ (1)

Letting m represent the mass of the turtle, apply aF

rrm=∑ to the turtle:

xx mafF =−=∑ s (2) and

yy maFFF =−=∑ gn (3)

Solving equation (2) for fs yields:

xmaf −=s

Because ay = 0 and Fg = mg, equation (3) becomes:

mgFF == gn

Substituting for fs and Fn in equation (1) and simplifying yields:

ga

mgma xx −

=−

=sμ (4)

The acceleration of the truck and turtle is given by:

tvv

tva xx

x ΔΔΔ i,f, −

==

or, because vf,x = 0,

tv

a xx Δ

i,−=

Substitute for ax in equation (4) to obtain:

tgv x

Δi,

s =μ


( )( ) 21.0s 12m/s 9.81

s 3600h 1

mim 1609

hmi 55

2s =××

=μ


421

47 •• [SSM] A 150-g block is projected up a ramp with an initial speed of 7.0 m/s. The coefficient of kinetic friction between the ramp and the block is 0.23. (a) If the ramp is inclined 25° with the horizontal, how far along the surface of the ramp does the block slide before coming to a stop? (b) The block then slides back down the ramp. What is the coefficient of static friction between the block and the ramp if the block is not to slide back down the ramp? Picture the Problem The force diagram shows the forces acting on the block as it slides up the ramp. Note that the block is accelerated by

kfr

and the x component of .gFr

We can use a constant-acceleration equation to express the displacement of the block up the ramp as a function of its acceleration and Newton’s second law to find the acceleration of the block as it slides up the ramp.

gFr

nFr

θ

x

y

kfr

m

(a) Use a constant-acceleration equation to relate the distance the block slides up the incline to its initial speed and acceleration:

xavv xxx Δ220

2 += or, because vx = 0,

xav xx Δ20 20 += ⇒

x

x

avx

2Δ

20−

= (1)

Apply aF

rrm=∑ to the block:

xx maFfF =−−=∑ θsingk (2)


Substituting nkk Ff μ= and mgF =g in equations (2) and (3) yields:

xmamgF =−− θμ sinnk (4) and


Eliminate Fn between equations (4) and (5) to obtain:

xmamgmg =−− θθμ sincosk

Solving for ax yields:

( )gax θθμ sincosk +−=

Substitute for a in equation (1) to obtain:

( )gvx x

θθμ sincos2Δ

k

20

+=

Chapter 5

422

Substitute numerical values and evaluate Δx:

( )( )[ ] ( ) m 0.4m 957.3

m/s 81.925sin25cos23.02m/s 0.7Δ 2

2

==°+°

=x

(b) At the point at which the block is instantaneously at rest, static friction becomes operative and, if the static friction coefficient is too high, the block will not resume motion, but will remain at the high point. We can determine the maximum possible value of μs for which the block still slides back down the incline by considering the equality of the static friction force and the component of the weight of the block down the ramp.

gFr

nFr

θ

x

y

max,sfrm

Apply aFrr

m=∑ to the block when it is in equilibrium at the point at which it is momentarily at rest:

0singmaxs, =−=∑ θFfFx (5) and

0cosgn =−=∑ θFFFy (6)

Solving equation (6) for Fn yields:

θcosgn FF =

Because nsmaxs, Ff μ= , equation (5) becomes:

0sincos ggs =− θθμ FF or

0sincoss =− θθμ ⇒ θμ tans =

Substitute the numerical value of θ and evaluate μs:

47.025tans =°=μ

48 •• An automobile is going up a 15º grade at a speed of 30 m/s. The coefficient of static friction between the tires and the road is 0.70. (a) What minimum distance does it take to stop the car? (b) What minimum distance would it take to stop if the car were going down the grade? Picture the Problem We can find the stopping distances by applying Newton’s second law to the automobile and then using a constant-acceleration equation. The friction force the road exerts on the tires and the component of the car’s weight along the incline combine to provide the net force that stops the car. The pictorial representation summarizes what we know about the motion of the car. We can use Newton’s second law to determine the acceleration of the car and a constant-acceleration equation to obtain its stopping distance.


423

x

θ00 =t

?1 =t

00 =xm/s 300 =v

01 =vmin1 xx =

(a) Using a constant-acceleration equation, relate the final speed of the car to its initial speed, acceleration, and displacement; solve for its displacement:

minmax,20

21 2 xavv xxx +=

or, because v1x = 0,

x

x

avxmax,

20

min 2−

= (1)

Draw the free-body diagram for the car going up the incline:

gFr

nFr

θ

x

y

maxs,fr

Noting that ,g mgF = apply

∑ = aFrr

m to the car: xx mamgfF =−−=∑ θsinmaxs, (2)

and 0cosn =−=∑ θmgFFy (3)

Substitute nsmax s, Ff μ= and Fn from

equation (3) in equation (2) and solve for xamax, :

( )θθμ sincossmax, +−= ga x

Substituting for xamax, in equation (1)

yields: ( )θθμ sincos2 s

20

min +=

gvx x

Substitute numerical values and evaluate xmin:

( )( ) ( )( ) m 49

15sin15cos70.0m/s 81.92m/s 30

2

2

min =°+°

=x

Chapter 5

424

(b) When the car is going down the incline, the static friction force is up the incline as shown in the free-body diagram to the right. Note the change in coordinate system from Part (a).

gFr

nFr

θx

y

maxs,fr

Apply ∑ = aF

rrm to the car: xx mafmgF =−=∑ maxs,sinθ

and 0cosn =−=∑ θmgFFy

Proceed as in (a) to obtain: ( )θμθ cossin smax, −= ga x

Substituting for xamax, in equation (1)

yields: ( )θμθ cossin2 s

20

min −−

=g

vx x

Substitute numerical values and evaluate xmin:

( )( ) ( )( ) km .110

15cos70.015sinm/s 81.92m/s 30

2

2

min =°−°

−=x

49 •• A rear-wheel-drive car supports 40 percent of its weight on its two drive wheels and has a coefficient of static friction of 0.70 with a horizontal straight road. (a) Find the vehicle’s maximum acceleration. (b) What is the shortest possible time in which this car can achieve a speed of 100 km/h? (Assume the engine has unlimited power.) Picture the Problem The friction force the road exerts on the tires provides the net force that accelerates the car. The pictorial representation summarizes what we know about the motion of the car. We can use Newton’s second law to determine the acceleration of the car and a constant-acceleration equation to calculate how long it takes it to reach 100 km/h.

x

00 =t ?1 =t00 =x

0 1

?1 =xkm/h 1001 =v00 =v


425

(a) Because 40% of the car’s weight is on its two drive wheels and the accelerating friction forces act just on these wheels, the free-body diagram shows just the forces acting on the drive wheels.

x

y

nFr

sfr

gmr

4.0

Apply ∑ = aFrr

m to the drive

wheels: xx mafF max,maxs, ==∑ (1)

and 04.0n =−=∑ mgFFy (2)

Use the definition of max s,f in

equation (1) and eliminate Fn between the two equations to obtain:

ga x smax, 4.0 μ=

Substitute numerical values and evaluate xmax,a :

( )( )22

2max,

m/s7.2m/s 747.2

m/s81.970.04.0

==

=xa

(b) Using a constant-acceleration equation, relate the initial and final velocities of the car to its acceleration and the elapsed time; solve for the time:

tavv xxx Δmax,,0,1 +=

or, because v0,x = 0 and Δt = t1,

x

x

av

tmax,

,11 =

Substitute numerical values and evaluate t1:

s10m/s2.747

kmm 1000

s 3600h 1

hkm100

21 =××

=t

50 •• You and your best pal make a friendly bet that you can place a 2.0-kg box against the side of a cart, as in Figure 5-63, and that the box will not fall to the ground, even though you guarantee to use no hooks, ropes, fasteners, magnets, glue, or adhesives of any kind. When your friend accepts the bet, you begin pushing the cart in the direction shown in the figure. The coefficient of static friction between the box and the cart is 0.60. (a) Find the minimum acceleration for which you will win the bet. (b) What is the magnitude of the frictional force in this case? (c) Find the force of friction on the box if the acceleration is twice the minimum needed for the box not to fall. (d) Show that, for a box of any mass, the box will not fall if the magnitude of the forward acceleration is a ≥ g/μs, where μs is the coefficient of static friction.

Chapter 5

426

Picture the Problem To hold the box in place, the acceleration of the cart and box must be great enough so that the static friction force acting on the box will equal the weight of the box. We can use Newton’s second law to determine the minimum acceleration required.

x

y

nFr

gFr

maxs,fr

(a) Noting that ,g mgF = apply

∑ = aFrr

m to the box:

xx maFF min,n ==∑ (1)

and 0maxs, =−=∑ mgfFy (2)

Substituting ns Fμ for max s,f in

equation (2) yields:

0ns =−mgFμ

Substitute for Fn from equation (1) to obtain:

( ) 0xmin,s =−mgmaμ ⇒s

min, μga x =

Substitute numerical values and evaluate xa min, :

22

min, m/s160.60

m/s81.9==xa

(b) From equation (2) we have: mgf =maxs,

Substitute numerical values and evaluate max s,f :

( )( ) N 20m/s 81.9kg 0.2 2maxs, ==f

(c) If a is twice that required to hold the box in place, fs will still have its maximum value given by:

N20maxs, =f

(d) Because sxmin, μga = , the box will not fall if .sμga ≥ 51 •• Two blocks attached by a string (Figure 5-64) slide down a 10º incline. Block 1 has mass m1 = 0.80 kg and block 2 has mass m2 = 0.25 kg. In addition, the kinetic coefficients of friction between the blocks and the incline are 0.30 for block 1 and 0.20 for block 2. Find (a) the magnitude of the acceleration of the blocks, and (b) the tension in the string.


427

Picture the Problem Assume that the string is massless and does not stretch. Then the blocks have a common acceleration and the tension in the string acts on both blocks in accordance with Newton’s third law of motion. Let down the incline be the +x direction. Draw the free-body diagrams for each block and apply Newton’s second law of motion and the definition of the kinetic friction force to each block to obtain simultaneous equations in ax and T. Draw the free-body diagram for the block whose mass is m1:

θT

m g

F

f

n, 1

1

k, 1

m1

x

y

Apply ∑ = aFrr

m to the upper block:

amgmTfFx

1

1k,1 sin=

++−=∑ θ (1)

and 0cos1n,1 =−=∑ θgmFFy (2)

The relationship between fk,1 and Fn,1 is:

n,1k,1k,1 Ff μ= (3)

Eliminate fk,1 and Fn,1 between (1), (2), and (3) to obtain:

amgmTgm

1

11k,1 sincos=

++− θθμ (4)

Draw the free-body diagram for the block whose mass is m2:

θ

n, 2

2

k, 2

F

m g2

2f

T

m

y

x

Chapter 5

428

Apply ∑ = aF

rrm to the block:

amgmTfFx

2

2k,2 sin=

+−−=∑ θ (5)

and 0cos2n,2 =−=∑ θgmFFy (6)

The relationship between fk,2 and Fn,2 is:

n,2k,2k,2 Ff μ= (7)

Eliminate fk,2 and Fn,2 between (5), (6), and (7) to obtain:

amgmTgm

1

22k,2 sincos=

+−− θθμ (8)

Add equations (4) and (8) to eliminate T, and solve for a :

gmm

mma ⎥

⎦

⎤⎢⎣

⎡++

−= θμμ

θ cossin21

2k,21k,1

Substitute numerical values and evaluate a :

( )( ) ( )( ) ( )2

2

m/s 96.0

m/s 81.910coskg 80.0kg 25.0

kg 80.030.0kg 25.020.010sin

=

⎥⎦

⎤⎢⎣

⎡°

++

−°=a

(b) Eliminate a between equations (4) and (8) and solve for T = T1 = T2 to obtain:

( )21

k,1k,221 cosmm

gmmT

+−

=θμμ


( )( )( )( ) N 18.0kg 80.0kg 25.0

10cosm/s 81.920.030.0kg 80.0kg 25.0 2

=+

°−=T

52 •• Two blocks of masses m1 and m2 are sliding down an incline as shown in Figure 5-64. They are connected by a massless rod. The coefficients of kinetic friction between the block and the surface are μ1 for block 1 and μ2 for block 2. (a) Determine the acceleration of the two blocks. (b) Determine the force that the rod exerts on each of the two blocks. Show that the forces are both 0 when μ1 = μ2 and give a simple, nonmathematical argument why this is true. Picture the Problem The free-body diagrams show the forces acting on the two blocks as they slide down the incline. Down the incline has been chosen as the +x direction. T is the force transmitted by the rod; it can be either tensile (T > 0) or


429

compressive (T < 0). By applying Newton’s second law to these blocks, we can obtain equations in T and ax from which we can eliminate either by solving them simultaneously. Once we have expressed T, the role of the rod will become apparent.

x

y

θ

2Tr

n,2Fr

gmr

2

k,2fr

2m

y

θ

k,1fr

gmr

1

n,1Fr

1Tr

1m

x

(a) Apply ∑ = aF

rrm to block 1: xx amfgmTF 1k,111 sin =−+=∑ θ

and ∑ =−= 0cos1n,1 θgmFFy

Apply ∑ = aFrr

m to block 2: xx amfTgmF 2k,222 sin =−−=∑ θ and ∑ =−= 0cos2n,2 θgmFFy

Letting T1 = T2 = T, use the definition of the kinetic friction force to eliminate fk,1 and Fn,1 between the equations for block 1 and fk,2 and Fn,1 between the equations for block 2 to obtain:

θμθ cossin 1111 gmTgmam x −+= (1) and

θμθ cossin 2222 gmTgmam x −−= (2)

Add equations (1) and (2) to eliminate T and solve for ax: ⎟⎟

⎠

⎞⎜⎜⎝

⎛++

−= θμμθ cossin21

2211

mmmmgax

(b) Rewrite equations (1) and (2) by dividing both sides of (1) by m1 and both sides of (2) by m2 to obtain.

θμθ cossin 11

gmTgax −+= (3)

and

θμθ cossin 22

gmTgax −−= (4)

Subtracting (4) from (3) and rearranging yields: ( ) θμμ cos21

21

21 gmm

mmT −⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

Chapter 5

430

If μ1 = μ2, T = 0 and the blocks move down the incline with the same acceleration of g(sinθ − μcosθ). Inserting a stick between them can’t change this; therefore, the stick must exert no force on either block. 53 •• [SSM] A block of mass m rests on a horizontal table (Figure 5-65). The block is pulled by a massless rope with a force F

r at an angle θ. The

coefficient of static friction is 0.60. The minimum value of the force needed to move the block depends on the angle θ. (a) Discuss qualitatively how you would expect this force to depend on θ. (b) Compute the force for the angles θ = 0º, 10º, 20º, 30º, 40º, 50º, and 60º, and make a plot of F versus θ for mg = 400 N. From your plot, at what angle is it most efficient to apply the force to move the block? Picture the Problem The vertical component of F

rreduces the normal

force; hence, the static friction force between the surface and the block. The horizontal component is responsible for any tendency to move and equals the static friction force until it exceeds its maximum value. We can apply Newton’s second law to the box, under equilibrium conditions, to relate F to θ.

x

y

nFr

θsfr

Fr

gmFrr

=g

m

(a) The static-friction force opposes the motion of the object, and the maximum value of the static-friction force is proportional to the normal force FN. The normal force is equal to the weight minus the vertical component FV of the force F. Keeping the magnitude F constant while increasing θ from zero results in an increase in FV and a decrease in Fn; thus decreasing the maximum static-friction force fmax. The object will begin to move if the horizontal component FH of the force F exceeds fmax. An increase in θ results in a decrease in FH. As θ increases from 0, the decrease in FN is larger than the decrease in FH, so the object is more and more likely to slip. However, as θ approaches 90°, FH approaches zero and no movement will be initiated. If F is large enough and if θ increases from 0, then at some value of θ the block will start to move. (b) Apply ∑ = aF

rrm to the block:

0cos s =−=∑ fFFx θ (1)

and 0sinn =−+=∑ mgFFFy θ (2)

Assuming that fs = fs,max, eliminate fs and Fn between equations (1) and (2) and solve for F:

θμθμ

sincos s

s

+=

mgF


431

Use this function with mg = 400 N to generate the following table:

θ (deg) 0 10 20 30 40 50 60 F (N) 240 220 210 206 208 218 235

The following graph of F(θ) was plotted using a spreadsheet program.

205

210

215

220

225

230

235

240

0 10 20 30 40 50 60

θ (degrees)

F (N

)

From the graph, we can see that the minimum value for F occurs when θ ≈ 32°. Remarks: An alternative to manually plotting F as a function of θ or using a spreadsheet program is to use a graphing calculator to enter and graph the function. 54 •• Consider the block in Problem 53. Show that, in general, the following results hold for a block of mass m resting on a horizontal surface whose coefficient of static friction is μs. (a) If you want to apply the minimum possible force to move the block, you should apply it with the force pulling upward at an angle θmin = tan–1 μs, and (b) the minimum force necessary to start the block moving is ( )mgF 2

ssmin 1 μμ −= . (c) Once the block starts moving, if you want to apply the least possible force to keep it moving, should you keep the angle at which you are pulling the same? increase it? decrease it?

Chapter 5

432

Picture the Problem The free-body diagram shows the forces acting on the block. We can apply Newton’s second law, under equilibrium conditions, to relate F to θ and then set its derivative with respect to θ equal to zero to find the value of θ that minimizes F.

x

y

nFr

θsfr

Fr

gmFrr

=g

m

(a) Apply ∑ = aF

rrm to the block: 0cos s =−=∑ fFFx θ (1)

and 0sinn =−+=∑ mgFFFy θ (2)


θμθμ

sincos s

s

+=

mgF (3)

To find θmin, differentiate F with respect to θ and set the derivative equal to zero for extrema of the function:

( ) ( )( )

( )( )

( )( )

extremafor 0sincos

cossinsincos

sincos

sincos

sincos

2s

ss

2s

ss

2s

ss

=+

+−=

+

+−

+

+=

θμθθμθμ

θμθ

θμθθ

μ

θμθ

μθ

θμθ

θmg

ddmgmg

dd

ddF

Solving for θmin yields:

s1

min tan μθ −=

(b) Use the reference triangle shown below to substitute for cosθ and sinθ in equation (3):

mg

mg

mgF

2s

s

2s

2s

s

2s

ss2

s

smin

1

11

111

μ

μ

μ

μμ

μ

μμμ

μ

+=

+

+=

++

+

=


433

(c) The coefficient of kinetic friction is less than the coefficient of static friction. An analysis identical to the one above shows that the minimum force one should apply to keep the block moving should be applied at an angle given by

k1

min tan μθ −= . Therefore, once the block is moving, the coefficient of friction will decrease, so the angle can be decreased. 55 •• Answer the questions in Problem 54, but for a force F

r that pushes

down on the block at an angle θ below the horizontal. Picture the Problem The vertical component of F

rincreases the normal force and

the static friction force between the surface and the block. The horizontal component is responsible for any tendency to move and equals the static friction force until it exceeds its maximum value. We can apply Newton’s second law to the box, under equilibrium conditions, to relate F to θ. (a) As θ increases from zero, F increases the normal force exerted by the surface and the static friction force. As the horizontal component of F decreases with increasing θ, one would expect F to continue to increase. x

y

nFr

θsfr

gmFrr

=gFr

m

(b) Apply ∑ = aFrr

m to the block:

0cos s =−=∑ fFFx θ (1)

and 0sinn =−−=∑ mgFFFy θ (2)


θμθμ

sincos s

s

−=

mgF (3)

Use this function with mg = 400 N to generate the table shown below.

θ (deg) 0 10 20 30 40 50 60 F (N) 240 273 327 424 631 1310 very

large

Chapter 5

434

The following graph of F as a function of θ, plotted using a spreadsheet program, confirms our prediction that F continues to increase with θ.

0

200

400

600

800

1000

1200

1400

0 10 20 30 40 50

θ (degrees)

F (N

)

(a) From the graph we see that: °= 0minθ

(b) Evaluate equation (3) for θ = 0° to obtain:

mgmgF ss

s

0sin0cosμ

μμ

=°−°

=

(c) You should keep the angle at 0°.

Remarks: An alternative to the use of a spreadsheet program is to use a graphing calculator to graph the function. 56 •• A 100-kg mass is pulled along a frictionless surface by a horizontal force F

r such that its acceleration is a1 = 6.00 m/s2 (Figure 5-66). A 20.0-kg mass

slides along the top of the 100-kg mass and has an acceleration of a2 = 4.00 m/s2. (It thus slides backward relative to the 100-kg mass.) (a) What is the frictional force exerted by the 100-kg mass on the 20.0-kg mass? (b) What is the net force acting on the 100-kg mass? What is the force F? (c) After the 20.0-kg mass falls off the 100-kg mass, what is the acceleration of the 100-kg mass? (Assume that the force F does not change.) Picture the Problem The forces acting on each of these masses are shown in the free-body diagrams below. m1 represents the mass of the 20.0-kg mass and m2 that of the 100-kg mass. As described by Newton’s 3rd law, the normal reaction force Fn,1 and the friction force fk,1 (= fk,2) act on both masses but in opposite directions. Newton’s second law and the definition of kinetic friction forces can be used to determine the various forces and the acceleration called for in this problem.


435

(a) Draw a free-body diagram showing the forces acting on the block whose mass is 20 kg:

x

y

n,1Fr

gmFrr

1g =

k,1fr1m

Apply ∑ = aFrr

m to this mass: xx amfF 11k,1 ==∑ (1)

and 01n,1 =−=∑ gmFFy (2)

Substitute numerical values in equation (1) and evaluate fk,1:

( )( ) N 0.80m/s 00.4kg 0.20 2k,1 ==f

(b) Draw a free-body diagram showing the forces acting on the block whose mass is 100 kg:

x

y

gmFrr

2g,2 =

k,2fr

n,2Fr

n,1Fr

Fr

2m

Apply ∑ = xx maF to the 100-kg

object and evaluate Fnet:

( )( )N600

m/s00.6kg100 222net

=

== xamF

Express F in terms of Fnet and fk,2: k,2net fFF +=

Substitute numerical values and evaluate F:

N 680N 80.0N 600 =+=F

(c) When the 20.0-kg object falls off, the 680-N force acts just on the 100-kg object and its acceleration is given by Newton’s second law:

2net m/s80.6kg100N680

===m

Fa

Chapter 5

436

57 •• A 60.0-kg block slides along the top of a 100-kg block. The 60.0-kg block has an acceleration of 3.0 m/s2 when a horizontal force of 320 N is applied, as in Figure 5-67. There is no friction between the 100-kg block and a horizontal frictionless surface, but there is friction between the two blocks. (a) Find the coefficient of kinetic friction between the blocks. (b) Find the acceleration of the 100-kg block during the time that the 60.0-kg block remains in contact. Picture the Problem The forces acting on each of these blocks are shown in the free-body diagrams to the right. m1 represents the mass of the 60-kg block and m2 that of the 100-kg block. As described by Newton’s 3rd law, the normal reaction force Fn,1 and the friction force fk,1 (= fk,2) act on both objects but in opposite directions. Newton’s second law and the definition of kinetic friction forces can be used to determine the coefficient of kinetic friction and acceleration of the 100-kg block.

(a) Apply ∑ = aF

rrm to the 60-kg

block: xx amfFF ,11k,1 =−=∑ (1)

and 01n,1 =−=∑ gmFFy (2)

Apply ∑ = xx maF to the 100-kg

block:

xamf ,22k,2 = (3)

Using equation (2), express the relationship between the kinetic friction forces 1,kf

rand 2,kf

r:

gmFfff 1kn,1kkk,2k,1 μμ ==== (4)

Substitute for fk,1 in equation (1) and solve for μ k: gm

amF

1

11k

−=μ

Substitute numerical values and evaluate μ k:

( )( )( )( )24.0

238.0m/s9.81kg60

m/s3.0kg60N3202

2

k

=

=−

=μ


437

(b) Substitute for fk,2 in equation (3) and solve for a2 to obtain: 2

1k,2 m

gma xμ

=

Substitute numerical values and evaluate a2,x:

( )( )( )

2

2

,2

m/s4.1

kg100m/s9.81kg600.238

=

=xa

58 •• The coefficient of static friction between a rubber tire and the road surface is 0.85. What is the maximum acceleration of a 1000-kg four-wheel-drive truck if the road makes an angle of 12º with the horizontal and the truck is (a) climbing and (b) descending? Picture the Problem Choose a coordinate system in which the +x direction is up the incline and apply Newton’s second law of motion. The free-body diagram shows the truck climbing the incline with maximum acceleration.

nFr

θ

Fr

g

maxs,fr

x

y

(a) Apply ∑ = aF

rrm to the truck

when it is climbing the incline: xx mamgfF =−=∑ θsinmaxs, (1)


Solve equation (2) for Fn and use the definition maxs,f to obtain:

θμ cossmaxs, mgf = (3)

Substitute maxs,f in equation (1) and

solve for ax:

( )θθμ sincoss −= gax

Substitute numerical values and evaluate ax:

( ) ( )[ ]2

2

m/s1.6

12sin12cos85.0m/s81.9

=

°−°=xa

Chapter 5

438

(b) When the truck is descending the incline with maximum acceleration, the static friction force points down the incline; that is, its direction is reversed Apply ∑ = xx maF to the

truck under these conditions to obtain:

xmamgf =−− θsinmaxs, (4)

Substituting for maxs,f in equation

(4) and solving for ax gives:

( )θθμ sincoss +−= gax

Substitute numerical values and evaluate ax:

( ) ( )[ ]2

2

m/s10

12sin12cos85.0m/s81.9

−=

°+°−=xa

59 •• A 2.0-kg block sits on a 4.0-kg block that is on a frictionless table (Figure 5-68). The coefficients of friction between the blocks are μs = 0.30 and μk = 0.20. (a) What is the maximum horizontal force F that can be applied to the 4.0-kg block if the 2.0-kg block is not to slip? (b) If F has half this value, find the acceleration of each block and the force of friction acting on each block. (c) If F is twice the value found in (a), find the acceleration of each block.. Picture the Problem The forces acting on each of the blocks are shown in the free-body diagrams to the right. m2 represents the mass of the 2.0-kg block and m4 that of the 4.0-kg block. As described by Newton’s 3rd law, the normal reaction force Fn,2 and the friction force fs,2 (= fs,4) act on both objects but in opposite directions. Newton’s second law and the definition of the maximum static friction force can be used to determine the maximum force acting on the 4.0-kg block for which the 2.0-kg block does not slide.

x

y

x

y

Fr

gmr

2

gmr

4

n,2Fr

n,4Fr

s,4fr

s,2fr

2,nFr

−

(a) Apply ∑ = aF

rrm to the 2.0-kg

block: max,22maxs,2, amfFx ==∑ (1)

and 02n,2 =−=∑ gmFFy (2)


439

Apply ∑ = aF

rrm to the 4.0-kg

block: max,44maxs,2, amfFFx =−=∑ (3)

and 04n,2n,4 =−−=∑ gmFFFy (4)

Using equation (2), express the relationship between the static friction forces max,2,sf

rand max,4,sf

r:

gmff s 2smax,4,maxs,2, μ== (5)

Substitute for max,2,sfr

in equation (1) and solve for max2,a :

ga smax2, μ=

Solve equation (3) for F = Fmax and substitute for max2,a and max4,a

to obtain:

( ) gmmgmgmF

s24

2ss4max

μμμ

+=+=

Substitute numerical values and evaluate Fmax:

( )( )( )N18N66.17

m/s9.810.30kg2.0kg4.0 2max

==

+=F

(b) Use Newton’s second law to express the acceleration of the blocks moving as a unit:

21

max21

21 mmF

mmFax +

=+

=

Substitute numerical values and evaluate a:

( )

2

221

m/s5.1

m/s 472.1kg4.0kg2.0

N66.17

=

=+

=xa

Because the friction forces are an action-reaction pair, the friction force acting on each block is given by:

xamf 1s =


( )( ) N9.2m/s472.1kg0.2 2s ==f

Chapter 5

440

(c) If F = 2Fmax, then the 2.0-kg block slips on the 4.0-kg block and the friction force (now kinetic) is given by:

gmff 2kk μ==

Use∑ = xx maF to relate the

acceleration of the 2.0-kg block to the net force acting on it and solve for a2,x:

xamgmf ,221kk == μ

and ga x k,2 μ=


( )( )2

22,2

m/s0.2

m/s96.1m/s 81.920.0

=

==xa

Use∑ = xx maF to relate the

acceleration of the 4.0-kg block to the net force acting on it:

xamgmF ,442k =− μ

Solving for a4,x yields: 4

2k,4 m

gmFa xμ−

=


( ) ( )( )( ) 22

,4 m/s8.7kg4.0

m/s9.81kg2.00.20N17.662=

−=xa

60 •• A 10.0-kg block rests on a 5.0-kg bracket, as shown in Figure 5-69. The 5.0-kg bracket sits on a frictionless surface. The coefficients of friction between the 10.0-kg block and the bracket on which it rests are μs = 0.40 and μk = 0.30. (a) What is the maximum force F that can be applied if the 10.0-kg block is not to slide on the bracket? (b) What is the corresponding acceleration of the 5.0-kg bracket?


441

Picture the Problem The top diagram shows the forces action on the 10-kg block and the bottom diagram shows the forces acting on the 5.0-kg bracket. If the block is not to slide on the bracket, the maximum value for F

rmust

equal the maximum value of fs. This value for F

rwill produce the maximum

acceleration of block and bracket system. We can apply Newton’s second law and the definition of fs,max to first calculate the maximum acceleration and then the maximum value of F.

x

sfr

n,5Fr

Fr

2

gmr

5

x

y

Fr

gmr

10

10n,Fr

sfr

n,10Fr

−

Apply ∑ = aF

rrm to the 10-kg

block when it is experiencing its maximum acceleration:

max,1010maxs, amFfFx =−=∑ (1)

and 010n,10 =−=∑ gmFFy (2)

Express the static friction force acting on the 10-kg block:

n,10smaxs, Ff μ= (3)

Eliminate maxs,f and Fn,10 from equations (1), (2) and (3) to obtain:

max,101010s amFgm =−μ (4)

Apply ∑ = xx maF to the bracket to

obtain:

max,5510s2 amgmF =− μ (5)

Because a5,max = a10,max, denote this acceleration by maxa . Eliminate F from equations (4) and (5) and solve for maxa :

105

10smax 2mm

gma+

=μ

(b) Substitute numerical values and evaluate maxa :

( )( )( )( )

22

2

max

m/s6.1m/s57.1

kg102kg5.0m/s9.81kg100.40

==

+=a

Solve equation (4) for maxFF = : ( )maxs10max1010s agmamgmF −=−= μμ

Chapter 5

442

(a) Substitute numerical values and evaluate F:

( ) ( )( )[ ] N24m/s1.57m/s9.810.40kg10 22 =−=F 61 ••• You and your friends push a 75.0-kg greased pig up an aluminum slide at the county fair, starting from the low end of the slide. The coefficient of kinetic friction between the pig and the slide is 0.070. (a) All of you pushing together (parallel to the incline) manage to accelerate the pig from rest at the constant rate of 5.0 m/s2 over a distance of 1.5 m, at which point you release the pig. The pig continues up the slide, reaching a maximum vertical height above its release point of 45 cm. What is the angle of inclination of the slide? (b) At the maximum height the pig turns around and begins to slip down once slide, how fast is it moving when it arrives at the low end of the slide? Picture the Problem The free-body diagram shows the forces acting on the pig sometime after you and your friends have stopped pushing on it but before it has momentarily stopped moving up the slide. We can use constant-acceleration equations and Newton’s second law to find the angle of inclination of the slide and the pig’s speed when it returns to bottom of the slide. The pictorial representation assigns a coordinate system and names to the variables that we need to consider in solving this problem.

θ

kfr

nFr

gFr

y

x

00 =xm 5.11 =x

2x02 =v

1v

00 =v

3v

02 =v

Pig going up the incline

Pig going down the incline

(a) Apply aF

rrm=∑ to the pig: xx maFfF =−−=∑ θsingk (1)


Substitute nkk Ff μ= and mgF =g in equation (1) to obtain:

xmamgF =−− θμ sinnk (3)

Solving equation (2) for Fn yields: θcosgn FF =

Substitute for Fn in equation (3) to obtain:

xmamgmg =−− θθμ sincosk

Solving for ax yields: ( )θθμ sincosk +−= gax (4)


443

Use a constant-acceleration equation to relate the distance d the pig slides up the ramp to its speed after you and your friends have stopped pushing, its final speed, and its acceleration:

davv x221

22 +=

or, because v2 = 0, dav x20 2

1 +=

Substituting for ax from equation (4) yields:

( )dgv θθμ sincos20 k21 +−= (5)

Use a constant-acceleration equation to relate the pig’s speed v1 after being accelerated to the distance Δx over which it was accelerated:

xavv xΔ220

21 +=

or, because v0 = 0, xav xΔ22

1 =

Substitute for 21v in equation (5) to

obtain: ( )dgxax θθμ sincos2Δ20 k +−= (6)

Use trigonometry to relate the vertical distance h risen by the pig during its slide to the distance d it moves up the slide:

θsindh = ⇒θsin

hd =

Substituting for d in equation (6) gives:

( )θ

θθμsin

sincos2Δ20 khgxax +−=

Solve for θ to obtain:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−= −

hg

xah

xΔtan k1 μθ

Substitute numerical values and evaluate θ:

( )( )( )( )

°=°=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−= −

7.5719.5

m 45.0m/s 9.81

m 5.1m/s 5.0m 45.0070.0tan

2

21θ

(b) Use a constant-acceleration equation to express v3 as a function of the pig’s acceleration down the incline adown, its initial speed v2, the distance Δd it slides down the incline:

davv Δ2 down22

23 +=

or, because v2 = 0, dav Δ2 down

23 = ⇒ dav Δ2 down3 = (7)

Chapter 5

444

When the pig is sliding down the incline, the kinetic friction force shown in the free-body diagram is reversed (it points up the incline … opposing the pig’s motion) and the pig’s acceleration is:

( )θμθ cossin kdown −= ga

Substitute for downa in equation (7) to obtain:

( ) dgv Δcossin2 k3 θμθ −= (8)

Δd is the sum of the distances the pig was pushed and then slid to a momentary stop:

θsinm 5.1Δ dd +=

Substituting for Δd in equation (8) yields:

( ) ⎟⎠⎞

⎜⎝⎛ +−=

θθμθ

sinm 5.1cossin2 k3

dgv

Substitute numerical values and evaluate v3:

( ) ( )[ ] m/s 9.1719.5sin

m 0.45m 5.1719.5cos070.0719.5sinm/s 81.92 23 =⎟

⎠⎞

⎜⎝⎛

°+°−°=v

62 ••• A 100-kg block on an inclined plane is attached to another block of mass m via a string, as in Figure 5-70. The coefficients of static and kinetic friction for the block and the incline are μs = 0.40 and μk = 0.20 and the plane is inclined 18º with horizontal. (a) Determine the range of values for m, the mass of the hanging block, for which the 100-kg block will not move unless disturbed, but if nudged, will slide down the incline. (b) Determine a range of values for m for which the 100-kg block will not move unless nudged, but if nudged will slide up the incline. Picture the Problem The free-body diagram shows the block sliding down the incline under the influence of a friction force, its weight, and the normal force exerted on it by the inclined surface. We can find the range of values for m for the two situations described in the problem statement by applying Newton’s second law of motion to, first, the conditions under which the block will not slide if pushed, and secondly, if pushed, the block will slide up the incline.


445

θ

F = Mg

f

x

y

Fn

k

g

M

(a) Assume that the block is sliding down the incline with a constant velocity and with no hanging weight (m = 0) and apply aF

rrm=∑ to the

block:

∑ =+−= 0sink θMgfFx and ∑ =−= 0cosn θMgFFy

Using nkk Ff μ= , eliminate Fn between the two equations and solve for the net force acting on the block:

θθμ sincosknet MgMgF +−=

If the block is moving, this net force must be nonnegative and:

( ) 0sincosk ≥+− Mgθθμ

This condition requires that: 325.018tantank =°=≤ θμ

Because °>= 18tan40.0kμ , the block will not move unless pushed. Because μk = 0.2, this condition is satisfied and:

0min =m

To find the maximum value, note that the maximum possible value for the tension in the rope is mg. For the block to move down the incline, the component of the block’s weight parallel to the incline minus the frictional force must be greater than or equal to the tension in the rope:

mgMgMg ≥− θμθ cossin k

Solving for maxm yields: ( )θμθ cossin kmax −≤ Mm

Substitute numerical values and evaluate maxm :

( ) ( )[ ]kg9.11

18cos20.018sinkg100max

=°−°≤m

Chapter 5

446

The range of values for m is: kg120 ≤≤ m

(b) If the block is being dragged up the incline, the frictional force will point down the incline, and:

gmMgMg mink cossin <+ θμθ

Solve for minm : ( )θμθ cossin kmin +> Mm

Substitute numerical values and evaluate minm :

( ) ( )[ ]kg9.49

18cos20.018sinkg100min

=°+°>m

If the block is not to move unless pushed:

gmMgMg maxs cossin >+ θμθ

Solve for maxm : ( )θμθ cossin smax +< Mm

Substitute numerical values and evaluate maxm :

( ) ( )[ ]kg9.68

18cos40.018sinkg100max

=°+°<m

The range of values for m is: kg69kg50 ≤≤ m 63 ••• A block of mass 0.50 kg rests on the inclined surface of a wedge of mass 2.0 kg, as in Figure 5-71. The wedge is acted on by a horizontal applied force F

r and slides on a frictionless surface. (a) If the coefficient of static friction

between the wedge and the block is μs = 0.80 and the wedge is inclined 35º with the horizontal, find the maximum and minimum values of the applied force for which the block does not slip. (b) Repeat part (a) with μs = 0.40. Picture the Problem The free-body diagram shows the forces acting on the 0.50-kg block when the acceleration is a minimum. Note the choice of coordinate system that is consistent with the direction of F

r.

Apply Newton’s second law to the block and to the block-plus-incline and solve the resulting equations for Fmin and Fmax.

sfr

nFr

x

y

θ

gmFrr

=g

m


447

(a) The maximum and minimum values of the applied force for which the block does not slip are related to the maximum and minimum accelerations of the block:

max,totmax xamF = (1)

and min,totmin xamF = (2)

where mtot is the combined mass of the block and incline.

Apply ∑ = aFrr

m to the 0.50-kg

block: xx mafFF =−=∑ θθ cossin sn (3)

and 0sincos sn =−+=∑ mgfFFy θθ (4)

Under minimum acceleration,

max s,s ff = . Express the relationship between max s,f and Fn:

nsmaxs, Ff μ=

Substitute max s,f for fs in equation

(4) and solve for Fn: θμθ sincos sn +=

mgF

Substitute for Fn and max s,f in equation (3) and solve for min,xx aa = :

θμθθμθ

sincoscossin

s

smin, +

−= gax

Substitute for ax, min in equation (2) to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

=θμθθμθ

sincoscossin

s

stotmin gmF

Substitute numerical values and evaluate minF :

( )( ) ( )( ) N 6.1

sin350.80cos35cos350.80sin35m/s9.81kg 5.2 2

min =⎥⎦

⎤⎢⎣

⎡°+°°−°

=F

To find the maximum acceleration, reverse the direction of sf

rand apply

∑ = aFrr

m to the block:

xx mafFF =+=∑ θθ cossin sn

and 0sincos sn =−−=∑ mgfFFy θθ

Proceed as above to obtain:

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=θμθθμθ

sincoscossin

s

smax, gax

Substitute for ax, max in equation (1) to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−+

=θμθθμθ

sincoscossin

s

stotmax gmF

Chapter 5

448

Substitute numerical values and evaluate maxF :

( )( ) ( )( ) N 84

sin350.80cos35cos350.80sin35m/s9.81kg 5.2 2

max =⎥⎦

⎤⎢⎣

⎡°−°°+°

=F

(b) Repeat Part (a) with μs = 0.40 to obtain:

N8.5min =F and N37max =F

64 ••• In your physics lab, you and your lab partners push a block of wood with a mass of 10.0 kg (starting from rest), with a constant horizontal force of 70 N across a wooden floor. In the previous week’s laboratory meeting, your group determined that the coefficient of kinetic friction was not exactly constant, but instead was found to vary with the object’s speed according to μk = 0.11/(1 + 2.3 × 10–4 v2)2. Write a spreadsheet program using Euler’s method to calculate and graph both the speed and the position of the block as a function of time from 0 to 10 s. Compare this result to the result you would get if you assumed the coefficient of kinetic friction had a constant value of 0.11. Picture the Problem The kinetic friction force fk is the product of the coefficient of sliding friction μk and the normal force Fn the surface exerts on the sliding object. By applying Newton’s second law in the vertical direction, we can see that, on a horizontal surface, the normal force is the weight of the sliding object. We can apply Newton’s second law in the horizontal (x) direction to relate the block’s acceleration to the net force acting on it. In the spreadsheet program, we’ll find the acceleration of the block from this net force (which is speed dependent), calculate the increase in the block’s speed from its acceleration and the elapsed time and add this increase to its speed at end of the previous time interval, determine how far it has moved in this time interval, and add this distance to its previous position to find its current position. We’ll also calculate the position of the block x2, under the assumption that μk = 0.11, using a constant-acceleration equation.


449

The spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows:

Cell Formula/Content Algebraic Form C9 C8+$B$6 tt Δ+ D9 D8+F9*$B$6 tav Δ+ E9 $B$5−($B$3)*($B$2)*$B$5/

(1+$B$4*D9^2)^2 ( )224k

1034.21 vmgF

−×+−

μ

F9 E10/$B$5 mF /net G9 G9+D10*$B$6 tvx Δ+ K9 0.5*5.922*I10^2 2

21 at

L9 J10-K10 2xx −

A B C D E F G H I J 1 g= 9.81 m/s2 2 Coeff1= 0.11 3 Coeff2= 2.30E−04 4 m= 10 kg 5 Fapp= 70 N 6 Δt= 0.05 s 7 8 t x x2 x−x2 9 t v Fnet a, x variableμ constantμ 10 0.00 0.00 0.00 0.00 0.00 0.00 0.00 11 0.05 0.30 59.22 5.92 0.01 0.05 0.01 0.01 0.01 12 0.10 0.59 59.22 5.92 0.04 0.10 0.04 0.03 0.01 13 0.15 0.89 59.22 5.92 0.09 0.15 0.09 0.07 0.02 14 0.20 1.18 59.22 5.92 0.15 0.20 0.15 0.12 0.03 15 0.25 1.48 59.23 5.92 0.22 0.25 0.22 0.19 0.04

205 9.75 61.06 66.84 6.68 292.37 9.75 292.37 281.48 10.89 206 9.80 61.40 66.88 6.69 295.44 9.80 295.44 284.37 11.07 207 9.85 61.73 66.91 6.69 298.53 9.85 298.53 287.28 11.25 208 9.90 62.07 66.94 6.69 301.63 9.90 301.63 290.21 11.42 209 9.95 62.40 66.97 6.70 304.75 9.95 304.75 293.15 11.61 210 10.00 62.74 67.00 6.70 307.89 10.00 307.89 296.10 11.79

The position of the block as a function of time, for a constant coefficient of friction (μk = 0.11) is shown as a solid line on the following graph and for a variable coefficient of friction, is shown as a dotted line. Because the coefficient of friction decreases with increasing particle speed, the particle travels slightly farther when the coefficient of friction is variable.

Chapter 5

450

0

50

100

150

200

250

300

0 2 4 6 8 10

t (s)

x (m

)

mu = variablemu = constant

The speed of the block as a function of time, with variable coefficient of kinetic friction, is shown in the following graph.

0

10

20

30

40

50

60

70

0 2 4 6 8 10t (s)

v (m

/s)

65 ••• In order to determine the coefficient of kinetic friction of a block of wood on a horizontal table surface, you are given the following assignment: Take the block of wood and give it an initial velocity across the surface of the table. Using a stopwatch, measure the time Δt it takes for the block to come to a stop and the total displacement Δx the block slides following the push. (a) Using Newton’s laws and a free-body diagram of the block, show that the expression for the coefficient of kinetic friction is given by μk = 2Δx/[(Δt)2g]. (b) If the block slides a distance of 1.37 m in 0.97 s, calculate μk. (c) What was the initial speed of the block?


451

Picture the Problem The free-body diagram shows the forces acting on the block as it moves to the right. The kinetic friction force will slow the block and, eventually, bring it to rest. We can relate the coefficient of kinetic friction to the stopping time and distance by applying Newton’s second law and then using constant-acceleration equations.

x

y

nFr

kfr

gmFrr

=g

m

(a) Apply ∑ = aF

rrm to the block of

wood: 0

and

n

k

=−=

=−=

∑

∑

mgFF

mafF

y

xx

Using the definition of fk, eliminate Fn between the two equations to obtain:

gax kμ−= (1)

Use a constant-acceleration equation to relate the acceleration of the block to its displacement and its stopping time:

( )221

0 ΔΔΔ tatvx xx += (2)

Relate the initial speed of the block, v0, to its displacement and stopping distance:

tvvtvx x Δ2

ΔΔ 0av ⎟

⎠⎞

⎜⎝⎛ +

==

or, because v = 0, ΔΔ 02

1 tvx x= (3)

Use this result to eliminate xv0 in equation (2):

( )221 ΔΔ tax x−= (4)

Substitute equation (1) in equation (4) and solve for μk: ( )2Δ

Δ2tgx

k =μ

(b) Substitute for Δx = 1.37 m and Δt = 0.97 s to obtain:

( )( )( )

30.0s0.97m/s9.81

m1.37222==kμ

(c) Use equation (3) to find v0: ( ) m/s8.2

s0.97m1.372

ΔΔ2

0 ===txv

Chapter 5

452

66 ••• (a) A block is sliding down an inclined plane. The coefficient of kinetic friction between the block and the plane is μk. Show that a graph of ax/cos θ versus tan θ (where ax is the acceleration down the incline and θ is the angle the plane is inclined with the horizontal) would be a straight line with slope g and intercept –μkg. (b) The following data show the acceleration of a block sliding down an inclined plane as a function of the angle θ that the plane is inclined with the horizontal:

θ (degrees) Acceleration (m/s2)25.0 1.69 27.0 2.10 29.0 2.41 31.0 2.89 33.0 3.18 35.0 3.49 37.0 3.79 39.0 4.15 41.0 4.33 43.0 4.72 45.0 5.11

Using a spreadsheet program, graph these data and fit a straight line (a Trendline in Excel parlance) to them to determine μk and g. What is the percentage difference between the obtained value of g and the commonly specified value of 9.81 m/s2? Picture the Problem The free-body diagram shows the forces acting on the block as it slides down an incline. We can apply Newton’s second law to these forces to obtain the acceleration of the block and then manipulate this expression algebraically to show that a graph of a/cosθ versus tanθ will be linear with a slope equal to the acceleration due to gravity and an intercept whose absolute value is the coefficient of kinetic friction.

nFr

θ x

y

kfr

gmFrr

=g

m

(a) Apply ∑ = aF

rrm to the block as

it slides down the incline:

∑

∑

=−=

=−=

0cosand

sin

n

k

θ

θ

mgFF

mafmgF

y

xx


453

Substitute μkFn for fk and eliminate Fn between the two equations to obtain:

( )θμθ cossin k−= gax

Divide both sides of this equation by cosθ to obtain: ktan

cosμθ

θggax −=

Note that this equation is of the form y = mx + b. Thus, if we graph a/cosθ versus tanθ, we should get a straight line with slope g and y-intercept −gμk. (b) A spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows:

Cell Formula/Content Algebraic FormC7 θ D7 a E7 TAN(C7*PI()/180)

⎟⎠⎞

⎜⎝⎛ ×

180tan πθ

F7 D7/COS(C7*PI()/180)

⎟⎠⎞

⎜⎝⎛ ×

180cos πθ

a

C D E F

6 θ a tanθ a/cosθ 7 25.0 1.69 0.466 1.866 8 27.0 2.10 0.510 2.362 9 29.0 2.41 0.554 2.751 10 31.0 2.89 0.601 3.370 11 33.0 3.18 0.649 3.786 12 35.0 3.49 0.700 4.259 13 37.0 3.78 0.754 4.735 14 39.0 4.15 0.810 5.338 15 41.0 4.33 0.869 5.732 16 43.0 4.72 0.933 6.451 17 45.0 5.11 1.000 7.220

Chapter 5

454

A graph of a/cosθ versus tanθ follows. From the curve fit (Excel’s Trendline

was used), 2m/s 77.9=g and 268.0m/s9.77m/s2.62

2

2

k ==μ .

y = 9.7681x − 2.6154

0

1

2

3

4

5

6

7

8

0.4 0.5 0.6 0.7 0.8 0.9 1.0

tanθ

a/c

osθ

The percentage error in g from the commonly accepted value of 9.81 m/s2 is:

%408.0m/s81.9

m/s77.9m/s81.9100 2

22

=⎟⎟⎠

⎞⎜⎜⎝

⎛ −

Drag Forces 67 • [SSM] A Ping-Pong ball has a mass of 2.3 g and a terminal speed of 9.0 m/s. The drag force is of the form bv2. What is the value of b? Picture the Problem The ping-pong ball experiences a downward gravitational force exerted by Earth and an upward drag force exerted by the air. We can apply Newton’s second law to the Ping-Pong ball to obtain its equation of motion. Applying terminal speed conditions will yield an expression for b that we can evaluate using the given numerical values. Let the downward direction be the +y direction. Apply yy maF =∑ to the Ping-Pong

ball:

ymabvmg =− 2

When the Ping-Pong ball reaches its terminal speed v = vt and ay = 0:

02t =−bvmg ⇒ 2

t

vmgb =


455

Substitute numerical values and evaluate b:

( )( )( )

kg/m108.2

m/s9.0m/s9.81kg102.3

4

2

23

−

−

×=

×=b

68 • A small pollution particle settles toward Earth in still air. The terminal speed is 0.30 mm/s, the mass of the particle is 1.0 × 10–10 g and the drag force is of the form bv. What is the value of b? Picture the Problem The pollution particle experiences a downward gravitational force exerted by Earth and an upward drag force exerted by the air. We can apply Newton’s second law to the particle to obtain its equation of motion. Applying terminal speed conditions will yield an expression for b that we can evaluate using the given numerical values. Let the downward direction by the +y direction. Apply yy maF =∑ to the particle: ymabvmg =−

When the particle reaches its terminal speed v = vt and ay = 0:

0t =−bvmg ⇒t

v

mgb =


( )( )

kg/s103.3

m/s103.0m/s81.9kg100.1

9

4

213

−

−

−

×=

××

=b

69 •• [SSM] A common classroom demonstration involves dropping basket-shaped coffee filters and measuring the time required for them to fall a given distance. A professor drops a single basket-shaped coffee filter from a height h above the floor, and records the time for the fall as Δt. How far will a stacked set of n identical filters fall during the same time interval Δt? Consider the filters to be so light that they instantaneously reach their terminal velocities. Assume a drag force that varies as the square of the speed and assume the filters are released oriented right-side up. Picture the Problem The force diagram shows n coffee filters experiencing an upward drag force exerted by the air that varies with the square of their terminal velocity and a downward gravitational force exerted by Earth. We can use the definition of average velocity and Newton’s second law to establish the dependence of the distance of fall on n. y

jnmgF ˆg =r

jnmgF ˆg =r

jCvF ˆ2td −=

r

Chapter 5

456

Express the distance fallen by 1 coffee filter, falling at its terminal speed, in time Δt:

tvd Δfilter 1 t,filter 1 = (1)

Express the distance fallen by n coffee filters, falling at their terminal speed, in time Δt:

tvd nn Δfilters t,filters = (2)

Divide equation (2) by equation (1) to obtain:

filter 1 t,

filters t,

filter 1 t,

filters t,

filter 1

filters

ΔΔ

vv

tvtv

dd nnn ==

Solving for dn filters yields: filter 1

filter 1 t,

filters t,filters d

vv

d nn ⎟

⎟⎠

⎞⎜⎜⎝

⎛= (3)

Apply yy maF =∑ to the coffee filters:

ymaCvnmg =− 2t

or, because ay = 0, 02

t =−Cvnmg

Solving for vt yields: n

Cmg

Cnmgv n ==filters t,

or, because ,filter 1 t, Cmgv =

nvv n filter 1 t,filters t, =

Substitute for vt, n filters in equation (3) to obtain:

filter 1filter 1filter 1 t,

filter 1 t,filters dnd

vnv

dn =⎟⎟⎠

⎞⎜⎜⎝

⎛=

This result tells us that n filters will fall farther, in the same amount of time, than 1 filter by a factor of n . 70 •• A skydiver of mass 60.0 kg can slow herself to a constant speed of 90 km/h by orienting her body horizontally, looking straight down with arms and legs extended. In this position she presents the maximum cross-sectional area and thus maximizes the air-drag force on her. (a) What is the magnitude of the drag force on the skydiver? (b) If the drag force is given by bv2, what is the value of b? (c) At some instant she quickly flips into a ″knife″ position, orienting her body vertically with her arms straight down. Suppose this reduces the value of b to 55 percent of the value in Parts (a) and (b). What is her acceleration at the instant she achieves the ″knife″ position?


457

Picture the Problem The skydiver experiences a downward gravitational force exerted by Earth and an upward drag force exerted by the air. Let the upward direction be the +y direction and apply Newton’s second law to the sky diver. (a) Apply yy maF =∑ to the sky

diver: ymamgF =−d (1)

or, because ay = 0, mgF =d (2)

Substitute numerical values and evaluate Fd:

( )( )N589

N6.588m/s81.9kg0.60 2d

=

==F

(b) Substitute Fd = b 2

tv in equation (2) to obtain:

mgbv =2t ⇒ 2

t

d2t v

Fvmgb ==


kg/m94.0

kg/m 9418.0

s 3600h 1

hkm 90

N588.62

=

=

⎟⎠⎞

⎜⎝⎛ ×

=b

(c) Solving equation (1) for ay yields: g

mFay −= d

Because b is reduced to 55 percent of its value in Part (b), so is Fd. Substitute numerical values and evaluate ay:

( )( )

downward ,m/s 41.4

m/s 81.9kg 0.60

N 6.58855.0

2

2

=

−=ya

71 •• Your team of test engineers is to release the parking brake so an 800-kg car will roll down a very long 6.0% grade in preparation for a crash test at the bottom of the incline. (On a 6.0 % grade the change in altitude is 6.0% of the horizontal distance traveled.) The total resistive force (air drag plus rolling friction) for this car has been previously established to be Fd = 100 N + (1.2 N⋅s2/m2)v2, where m and v are the mass and speed of the car. What is the terminal speed for the car rolling down this grade?

Chapter 5

458

Picture the Problem The free-body diagram shows the forces acting on the car as it descends the grade with its terminal speed and a convenient coordinate system. The application of Newton’s second law with a = 0 and Fd equal to the given function will allow us to solve for the terminal speed of the car.

nFr

θx

y

dFr

gmFrr

=g

Apply ∑ = xx maF to the car:

xmaFmg =− dsinθ

or, because v = vt and ax = 0, 0sin d =− Fmg θ

Substitute for Fd to obtain: ( ) 0m/sN1.2N100sin 2

t22 =⋅−− vmg θ

Solving for vt yields:

22t m/sN1.2N100sin

⋅−

=θmgv


( )( ) m/s25m/sN1.2

N1000.6sinm/s81.9kg80022

2

t =⋅

−°=v

72 ••• Small slowly moving spherical particles experience a drag force given by Stokes’ law: Fd = 6πηrv, where r is the radius of the particle, v is its speed, and η is the coefficient of viscosity of the fluid medium. (a) Estimate the terminal speed of a spherical pollution particle of radius 1.00 × 10–5 m and density of 2000 kg/m3. (b) Assuming that the air is still and that η is 1.80 × 10–5 N⋅s/m2, estimate the time it takes for such a particle to fall from a height of 100 m. Picture the Problem Let the downward direction be the +y direction and apply Newton’s second law to the particle to obtain an equation from which we can find the particle’s terminal speed. (a) Apply yy maF =∑ to a pollution

particle: ymarvmg =− πη6

or, because ay = 0,

06 t =− rvmg πη ⇒r

mgvπη6t =


459

Express the mass of a sphere in terms of its volume: ⎟⎟

⎠

⎞⎜⎜⎝

⎛==

34 3rVm πρρ

Substitute for m to obtain:

ηρ

92 2

tgrv =


( ) ( )( )( ) cm/s42.2cm/s 422.2

s/mN101.809m/s9.81kg/m2000m101.002

25

2325

t ==⋅×

×= −

−

v

(b) Use distance equals average speed times the fall time to find the time to fall 100 m at 2.42 cm/s:

h15.1s10128.4cm/s2.422cm10 3

4

=×==t

73 ••• [SSM] You are on an environmental chemistry internship, and are in charge of a sample of air that contains pollution particles of the size and density given in Problem 72. You capture the sample in an 8.0-cm-long test tube. You then place the test tube in a centrifuge with the midpoint of the test tube 12 cm from the center of the centrifuge. You set the centrifuge to spin at 800 revolutions per minute. (a) Estimate the time you have to wait so that nearly all of the pollution particles settle to the end of the test tube. (b) Compare this to the time required for a pollution particle to fall 8.0 cm under the action of gravity and subject to the drag force given in Problem 72. Picture the Problem The motion of the centrifuge will cause the pollution particles to migrate to the end of the test tube. We can apply Newton’s second law and Stokes’ law to derive an expression for the terminal speed of the sedimentation particles. We can then use this terminal speed to calculate the sedimentation time. We’ll use the 12 cm distance from the center of the centrifuge as the average radius of the pollution particles as they settle in the test tube. Let R represent the radius of a particle and r the radius of the particle’s circular path in the centrifuge. (a) Express the sedimentation time in terms of the sedimentation speed vt:

tsediment v

xt Δ=Δ (1)

Apply ∑ = radialradial maF to a

pollution particle:

ct6 maRv =πη

Chapter 5

460

Express the mass of the particle in terms of its radius R and density ρ:

ρπρ 334 RVm ==

Express the acceleration of the pollution particles due to the motion of the centrifuge in terms of their orbital radius r and period T:

2

2

2

2

c4

2

Tr

rT

r

rva π

π

=⎟⎠⎞

⎜⎝⎛

==

Substitute for m and ac and simplify to obtain: 2

33

2

23

34

t 31646

TrR

TrRRv ρππρππη =⎟⎟⎠

⎞⎜⎜⎝

⎛=

Solving for tv yields:

2

22

t 98

TrRv

ηρπ

=

Substitute for tv in equation (1) and simplify to obtain:

22

2

2

22sediment 89

98 rR

xT

TrR

xtρπ

η

ηρπ

Δ=

Δ=Δ

Substitute numerical values and evaluate sedimenttΔ :

( )

( )( )ms 38ms 47.38

m10m12.0mkg20008

cm 0.8

s 60min 1

minrev800

1m

sN101.89

253

2

2

25

sediment ≈=⎟⎠⎞

⎜⎝⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⋅⎟⎠⎞

⎜⎝⎛ ⋅

×

=Δ−

−

πt

(b) In Problem 72 it was shown that the rate of fall of the particles in air is 2.42 cm/s. Find the time required to fall 8.0 cm in air under the influence of gravity:

s 31.3cm/s42.2cm8.0ΔΔ air ===

vxt

Find the ratio of the two times: 86ms38.47s 31.3

ΔΔ

sediment

air ≈=t

t

With the drag force in Problem 72 it takes about 86 times longer than it does using the centrifuge.


461

Motion Along a Curved Path 74 • A rigid rod with a 0.050-kg ball at one end rotates about the other end so the ball travels at constant speed in a vertical circle with a radius of 0.20 m. What is the maximum speed of the ball so that the force of the rod on the ball does not exceed 10 N? Picture the Problem The force of the rod on the ball is a maximum when the ball is at the bottom of the vertical circle. We can use Newton’s second law to relate the force of the rod on the ball to the mass m of the ball, the radius r of its path, and the constant speed of the ball along its circular path. The diagram to the right shows the forces acting on the ball when it is at the bottom of the vertical circle.

gmFrr

=g

maxTr

r

Apply radialradial maF =∑ to the ball: r

vmmgT2max

max =−

Solving for vmax yields: rg

mTv ⎟

⎠⎞

⎜⎝⎛ −= max

max

Substitute numerical values and evaluate vmax:

( ) m/s 2.6m 20.0m/s 81.9kg 050.0

N 10 2max =⎟⎟

⎠

⎞⎜⎜⎝

⎛−=v

75 • [SSM] A 95-g stone is whirled in a horizontal circle on the end of an 85-cm-long string. The stone takes 1.2 s to make one complete revolution. Determine the angle that the string makes with the horizontal.

Chapter 5

462

Picture the Problem The only forces acting on the stone are the tension in the string and the gravitational force. The centripetal force required to maintain the circular motion is a component of the tension. We’ll solve the problem for the general case in which the angle with the horizontal is θ by applying Newton’s second law of motion to the forces acting on the stone. Apply ∑ = aF

rrm to the stone:

rvmmaTFx

2

ccos ===∑ θ (1)

and 0sin =−=∑ mgTFy θ (2)

Use the right triangle in the diagram to relate r, L, and θ :

θcosLr = (3)

Eliminate T and r between equations (1), (2) and (3) and solve for v2:

θθ coscot2 gLv = (4)

Express the speed of the stone in terms of its period:

rev1

2t

rv π= (5)

Eliminate v between equations (4) and (5) and solve for θ : ⎟

⎟⎠

⎞⎜⎜⎝

⎛= −

Lgt

2

2rev11

4sin

πθ

Substitute numerical values and evaluate θ :

( )( )( ) °=⎥

⎦

⎤⎢⎣

⎡= − 25

m0.854s1.2m/s9.81sin 2

221

πθ

76 •• A 0.20-kg stone is whirled in a horizontal circle on the end of an 0.80-m-long string. The string makes an angle of 20º with the horizontal. Determine the speed of the stone.


463

Picture the Problem The only forces acting on the stone as it moves in a horizontal circle are the tension in the string and the gravitational force. The centripetal force required to maintain the circular motion is a component of the tension. We’ll solve the problem for the general case in which the angle with the horizontal is θ by applying Newton’s second law of motion to the forces acting on the stone. Apply ∑ = aF

rrm to the stone:

rvmmaTFx

2

ccos ===∑ θ (1)

and 0sin =−=∑ mgTFy θ (2)

Use the right triangle in the diagram to relate r, L, and θ:

θcosLr = (3)

Eliminate T and r between equations (1), (2), and (3) and solve for v:

θθ coscotgLv =

Substitute numerical values and evaluate v:

( )( )m/s 5.4

20cos20cotm0.80m/s9.81 2

=

°°=v

77 •• A 0.75-kg stone attached to a string is whirled in a horizontal circle of radius 35 cm as in the tetherball Example 5-11. The string makes an angle of 30º with the vertical. (a) Find the speed of the stone. (b) Find the tension in the string. Picture the Problem The only forces acting on the stone are the tension in the string and the gravitational force. The centripetal force required to maintain the circular motion is a component of the tension. We’ll solve the problem for the general case in which the angle with the vertical is θ by applying Newton’s second law of motion to the forces acting on the stone.

Chapter 5

464

(a) Apply ∑ = aFrr

m to the stone: rvmmaTFx

2

csin ===∑ θ (1)

and 0cos =−=∑ mgTFy θ (2)

Eliminate T between equations (1) and (2) and solve for v:

θtanrgv =


( )( )m/s1.4

tan30m/s9.81m0.35 2

=

°=v

(b) Solving equation (2) for T yields:

θcosmgT =


( )( ) N5.8cos30

m/s9.81kg0.75 2

=°

=T

78 •• A pilot with a mass of 50 kg comes out of a vertical dive in a circular arc such that at the bottom of the arc her upward acceleration is 3.5g. (a) How does the magnitude of the force exerted by the airplane seat on the pilot at the bottom of the arc compare to her weight? (b) Use Newton’s laws of motion to explain why the pilot might be subject to a blackout. This means that an above normal volume of blood ″pools″ in her lower limbs. How would an inertial reference frame observer describe the cause of the blood pooling? Picture the Problem The diagram shows the forces acting on the pilot when her plane is at the lowest point of its dive. nF

ris the force the

airplane seat exerts on her. We’ll apply Newton’s second law for circular motion to determine Fn and the radius of the circular path followed by the airplane.

gmFrr

=g

r

nFr

(a) Apply radialradial maF =∑ to the

pilot:

cn mamgF =− ⇒ ( )cn agmF +=

Because ga 5.3c = : ( ) mgggmF 5.45.3n =+=


465

The ratio of Fn to her weight is:

5.45.4n ==mg

mgmgF

(b) An observer in an inertial reference frame would see the pilot’s blood continue to flow in a straight line tangent to the circle at the lowest point of the arc. The pilot accelerates upward away from this lowest point and therefore it appears, from the reference frame of the plane, as though the blood accelerates downward. 79 •• A 80.0-kg airplane pilot pulls out of a dive by following, at a constant speed of 180 km/h, the arc of a circle whose radius is 300 m. (a) At the bottom of the circle, what are the direction and magnitude of his acceleration? (b) What is the net force acting on him at the bottom of the circle? (c) What is the force exerted on the pilot by the airplane seat? Picture the Problem The diagram shows the forces acting on the pilot when his plane is at the lowest point of its dive. nF

ris the force the airplane seat

exerts on the pilot. We’ll use the definitions of centripetal acceleration and centripetal force and apply Newton’s second law to calculate these quantities and the normal force acting on the pilot. gmF

rr=g

r

nFr

(a) The pilot’s acceleration is centripetal and given by: upward ,

2

c rva =

Substitute numerical values and evaluate ac:

upward ,m/s 33.8

m/s 333.8m 300

s 3600h 1

hkm180

2

2

2

c

=

=⎟⎠⎞

⎜⎝⎛ ×

=a

(b) The net force acting on her at the bottom of the circle is the force responsible for her centripetal acceleration:

( )( )upward N,676

m/s333.8kg0.80 2cnet

=

== maF

(c) Apply ∑ = yy maF to the pilot: cn mamgF =− ⇒ ( )cn agmF +=

Chapter 5

466

Substitute numerical values and evaluate nF :

( )( )upward kN, 45.1

m/s 33.8m/s 81.9kg 0.80 22n

=

+=F

80 •• A small object of mass m1 moves in a circular path of radius r on a frictionless horizontal tabletop (Figure 5-72). It is attached to a string that passes through a small frictionless hole in the center of the table. A second object with a mass of m2 is attached to the other end of the string. Derive an expression for r in terms of m1, m2, and the time T for one revolution. Picture the Problem Assume that the string is massless and that it does not stretch. The free-body diagrams for the two objects are shown to the right. The hole in the table changes the direction the tension in the string (which provides the centripetal force required to keep the object moving in a circular path) acts. The application of Newton’s second law and the definition of centripetal force will lead us to an expression for r as a function of m1, m2, and the time T for one revolution.

m gm g

T

x

x

y

F

m m1

1

2

2

n

T

''

Apply ∑ = xx maF to both objects

and use the definition of centripetal acceleration to obtain:

0'2 =−Tgm and

rvmamT

2

1c1' ==

Eliminating T between these equations gives:

02

12 =−rvmgm (1)

Express the speed v of the object in terms of the distance it travels each revolution and the time T for one revolution:

Trv π2

=

Substitute for v in equation (1) to obtain: 04

2

2

12 =−T

rmgm π⇒

12

22

4 mgTmrπ

=


467

81 •• [SSM] A block of mass m1 is attached to a cord of length L1, which is fixed at one end. The block moves in a horizontal circle on a frictionless tabletop. A second block of mass m2 is attached to the first by a cord of length L2 and also moves in a circle on the same frictionless tabletop, as shown in Figure 5-73. If the period of the motion is T, find the tension in each cord in terms of the given symbols. Picture the Problem The free-body diagrams show the forces acting on each block. We can use Newton’s second law to relate these forces to each other and to the masses and accelerations of the blocks.

m gm g

x x

y

F

m m1

1 2

T2T 2

Fn,1n,2

y

T12

Apply ∑ = xx maF to the block

whose mass is m1: 1

21

121 LvmTT =−


whose mass is m2: 21

22

22 LLvmT+

=

Relate the speeds of each block to their common period T and their distance from the center of the circle:

( )T

LLvTLv 21

21

12and2 +

==ππ

In the second force equation, substitute for v2, and simplify to obtain:

( )[ ]2

21222

⎟⎠⎞

⎜⎝⎛+=

TLLmT π

Substitute for T2 and v1 in the first force equation to obtain: ( )[ ]

2

1121212

⎟⎠⎞

⎜⎝⎛++=

TLmLLmT π

82 •• A particle moves with constant speed in a circle of radius 4.0 cm. It takes 8.0 s to complete each revolution. (a) Draw the path of the particle to scale, and indicate the particle’s position at 1.0-s intervals. (b) Sketch the displacement vectors for each interval. These vectors also indicate the directions for the average-velocity vectors for each interval. (c) Graphically find the magnitude of the change in the average velocity Δvr for two consecutive 1-s intervals. Compare Δvr /Δt, measured in this way, with the magnitude of the instantaneous acceleration computed from ac = v2/r.

Chapter 5

468

Picture the Problem (a) and (b) The path of the particle and its position at 1-s intervals are shown in the following diagram. The displacement vectors are also shown. The velocity vectors for the average velocities in the first and second intervals are along 01r

r and ,12rr

respectively. Δvr points toward the center of the

circle.

ΔΔ

Δ

0=t

s 1=t

s 2=t

s 3=t

s 4=t

s 5=t

s 6=t

s 7=t

vr 01r

r

12rr

r

(c) Use the diagram below to show that:

°= 5.22sin2Δ rr

Express the magnitude of the average velocity of the particle along the chords:

tr

trv

Δ5.22sin2

ΔΔ

av°

==r

Using the diagram below, express Δv in terms of v1 (= v2):

1vr

2vr

vr

Δ

°45

°= 5.22sin2Δ 1vv

Express Δv using avv as v1:

tr

trv

Δ5.22sin4

5.22sinΔ

5.22sin22Δ

2 °=

°⎟⎠⎞

⎜⎝⎛ °

=


469

Express tva

ΔΔ

= :

( )2

2

2

Δ5.22sin4

ΔΔ

5.22sin4

tr

tt

r

a °=

°

=

Substitute numerical values and evaluate a:

( )( )

cm/s 3.2

cm/s 34.2s 0.1

5.22sincm 0.44

2

22

2

=

=°

=a

The radial acceleration of the particle is given by: r

va2

c =

Express the speed v (= v1 = v2 …) of the particle along its circular path: T

rv π2=

Substituting for v in the expression for ac yields:

2

2

2

c4

2

Tr

rT

r

a ππ

=⎟⎠⎞

⎜⎝⎛

=


( )( )

2

22

2

c

cm/s 5.2

cm/s 47.2s 0.8

cm 0.44

=

==πa

Compare ac and a by taking their ratio: 06.1

cm/s34.2cm/s47.2

2

2c ==

aa

⇒ aa 1.1c =

83 •• You are swinging your younger sister in a circle of radius 0.75 m, as shown in Figure 5-74. If her mass is 25 kg and you arrange it so she makes one revolution every 1.5 s, (a) what are the magnitude and direction of the force that must be exerted by you on her? (Model her as a point particle.) (b) What are the magnitude and direction of the force she exerts on you? Picture the Problem The following diagram shows the free-body diagram for the child superimposed on a pictorial representation of her motion. The force you exert on your sister is F

rand the angle it makes with respect to the direction we’ve

chosen as the positive y direction is θ. We can infer her speed from the given information concerning the radius of her path and the period of her motion. Applying Newton’s second law will allow us to find both the direction and magnitude of F

r.

Chapter 5

470

(a) Apply ∑ = aF

rrm to the child: ∑ ==

rvmFFx

2

sinθ

and ∑ =−= 0cos mgFFy θ

Eliminate F between these equations and solve for θ to obtain: ⎥

⎦

⎤⎢⎣

⎡= −

rgv2

1tanθ

Express v in terms of the radius and period of the child’s motion: T

rv π2=

Substitute for v in the expression for θ to obtain: ⎥

⎦

⎤⎢⎣

⎡= −

2

21 4tan

gTrπθ


( )( )( )

horizontal above 53

3.53s1.5m/s9.81

m0.754tan 22

21

°=

°=⎥⎦

⎤⎢⎣

⎡= − πθ

Solve the y equation for F:

θcosmgF =


( )( ) kN 41.0cos53.3

m/s9.81kg25 2

=°

=F

(b) The force your sister exerts on you is the reaction force to the force you exert on her. Thus its magnitude is the same as the force you exert on her (0.41 kN) and its direction is 53° below horizontal.


471

84 •• The string of a conical pendulum is 50.0 cm long and the mass of the bob is 0.25 kg. (a) Find the angle between the string and the horizontal when the tension in the string is six times the weight of the bob. (b) Under those conditions, what is the period of the pendulum? Picture the Problem The following diagram shows the free-body diagram for the bob of the conical pendulum superimposed on a pictorial representation of its motion. The tension in the string is F

rand the angle it makes with respect to the

direction we’ve chosen as the +x direction isθ. We can find θ from the y equation and the information provided about the tension. Then, by using the definition of the speed of the bob in its orbit and applying Newton’s second law as it describes circular motion, we can find the period T of the motion.

(a) Apply ∑ = aF

rrm to the

pendulum bob: ∑ ==

rvmFFx

2

cosθ

and ∑ =−= 0sin mgFFy θ

Using the given information that F = 6mg, solve the y equation for θ and simplify to obtain:

°=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛= −− 6.9

6sinsin 11

mgmg

Fmgθ

(b) With F = 6mg, solve the x equation for v:

θcos6rgv =

Relate the period T of the motion to the speed of the bob and the radius of the circle in which it moves:

θππcos6

22rg

rv

rT ==

From the diagram, one can see that:

θcosLr =

Chapter 5

472

Substitute for r in the expression for the period and simplify to obtain: g

LT6

2π=

Substitute numerical values and evaluate T: ( ) s 58.0

m/s9.816m 0.502 2 == πT

85 •• A 100-g coin sits on a horizontally rotating turntable. The turntable makes exactly 1.00 revolution each second. The coin is located 10 cm from the axis of rotation of the turntable. (a) What is the frictional force acting on the coin? (b) If the coin slides off the turntable when it is located more than 16.0 cm from the axis of rotation, what is the coefficient of static friction between the coin and the turntable? Picture the Problem The static friction force fs is responsible for keeping the coin from sliding on the turntable. Using Newton’s second law of motion, the definition of the period of the coin’s motion, and the definition of the maximum static friction force, we can find the magnitude of the friction force and the value of the coefficient of static friction for the two surfaces.

x

y

nFr

sfr

gmFrr

=g

(a) Apply ∑ = aF

rrm to the coin: ∑ ==

rvmfF sx

2

(1)

and ∑ =−= 0n mgFFy

If T is the period of the coin’s motion, its speed is given by: T

rv π2=

Substitute for v in equation (1) and simplify to obtain: 2

24T

mrfsπ

=


( )( )( )

N 40.0

s 1.00m 0.10kg (0.1004π

2

2

=

=sf

(b) Determine Fn from the y equation: Fn = mg


473

If the coin is on the verge of sliding at r = 16 cm, fs = fs,max. Solve for μs in terms of fs,max and Fn: 2

22

2

n

max,s

44

gTr

mgT

mr

Ffs π

π

μ ===


( )( )( )

644.0s1.00m/s9.81

m0.1604π22

2

s ==μ

86 •• A 0.25-kg tether ball is attached to a vertical pole by a 1.2-m cord. Assume the radius of the ball is negligible. If the ball moves in a horizontal circle with the cord making an angle of 20º with the vertical, (a) what is the tension in the cord? (b) What is the speed of the ball? Picture the Problem The forces acting on the tetherball are shown superimposed on a pictorial representation of the motion. The horizontal component of T

ris the

centripetal force. Applying Newton’s second law of motion and solving the resulting equations will yield both the tension in the cord and the speed of the ball.

(a) Apply ∑ = aF

rrm to the

tetherball: ∑ =°=

rvmTFx

2

20sin

and ∑ =−°= 020cos mgTFy

Solve the y equation for T:

°=

20cosmgT


( )( ) N 6.220cos

m/s9.81kg0.25 2

=°

=T

(b) Eliminate T between the force equations and solve for v to obtain:

°= 20tanrgv

Note from the diagram that: °= 20sinLr

Substitute for r in the expression for v to obtain:

°°= 20tan20singLv

Chapter 5

474


( )( )m/s 2.1

20tan20sinm1.2m/s9.81 2

=

°°=v

87 ••• A small bead with a mass of 100 g (Figure 5-75) slides without friction along a semicircular wire with a radius of 10 cm that rotates about a vertical axis at a rate of 2.0 revolutions per second. Find the value of θ for which the bead will remain stationary relative to the rotating wire. Picture the Problem The semicircular wire of radius 10 cm limits the motion of the bead in the same manner as would a 10-cm string attached to the bead and fixed at the center of the semicircle. The horizontal component of the normal force the wire exerts on the bead is the centripetal force. The application of Newton’s second law, the definition of the speed of the bead in its orbit, and the relationship of the frequency of a circular motion to its period will yield the angle at which the bead will remain stationary relative to the rotating wire.

θ

θ

L

y

x r m

mg

Fn

Apply∑ = aF

rrm to the bead: ∑ ==

rvmFFx

2

n sinθ

and ∑ =−= 0cosn mgFFy θ

Eliminate Fn from the force equations to obtain: rg

v2

tan =θ

The frequency of the motion is the reciprocal of its period T. Express the speed of the bead as a function of the radius of its path and its period:

Trv π2

=

Using the diagram, relate r to L and θ :

θsinLr =


475

Substitute for r and v in the expression for tanθ and solve for θ : ⎥

⎦

⎤⎢⎣

⎡= −

LgT

2

21

4cos

πθ


( )( )( ) °=⎥

⎦

⎤⎢⎣

⎡= − 52

m0.104πs0.50m/s9.81cos 2

221θ

Centripetal Force 88 • A car speeds along the curved exit ramp of a freeway. The radius of the curve is 80.0 m. A 70.0-kg passenger holds the armrest of the car door with a 220-N force in order to keep from sliding across the front seat of the car. (Assume the exit ramp is not banked and ignore friction with the car seat.) What is the car’s speed? Picture the Problem The force F

r the

passenger exerts on the armrest of the car door is the radial force required to maintain the passenger’s speed around the curve and is related to that speed through Newton’s second law of motion.

x

y

nFr

Fr

gmFrr

=g

passenger

Apply ∑ = xx maF to the forces

acting on the passenger: rvmF

2

= ⇒mrFv =


( )( ) m/s 9.15kg70.0

N220m80.0==v

89 • [SSM] The radius of curvature of the track at the top of a loop-the-loop on a roller-coaster ride is 12.0 m. At the top of the loop, the force that the seat exerts on a passenger of mass m is 0.40mg. How fast is the roller-coaster car moving as it moves through the highest point of the loop. Picture the Problem There are two solutions to this problem. Pictorial diagrams that include a free-body diagram for each solution are shown below. The speed of the roller coaster is embedded in the expression for its radial acceleration and the radial acceleration is determined by the net radial force force exerted on the passenger by the seat of the roller coaster. We can use Newton’s second law to relate the net force on the passenger to the speed of the roller coaster.

Chapter 5

476

r

m

mg

mg0.40

vr

First solution

r

m

mg

mg0.40 vr

Second solution First solution Apply ∑ = radialradial maF to the

passenger:

rvmmgmg

2

40.0 =+ ⇒ grv 40.1=


( )( )( )m/s 8.12

m12.0m/s9.811.40 2

=

=v

Second solution Apply ∑ = radialradial maF to the

passenger:

rvmmgmg

2

40.0 =− ⇒ grv 60.0=


( )( )( )m/s .48

m12.0m/s9.810.60 2

=

=v

90 •• On a runway of a decommissioned airport, a 2000-kg car travels at a constant speed of 100 km/h. At 100-km/h the air drag on the car is 500 N. Assume that rolling friction is negligible. (a) What is the force of static friction exerted on the car by the runway surface, and what is the minimum coefficient of static friction necessary for the car to sustain this speed? (b) The car continues to travel at 100 km/h, but now along a path with radius of curvature r. For what value of r will the angle between the static frictional force vector and the velocity vector equal 45.0°, and for what value of r will it equal 88.0°? (c) What is the minimum coefficient of static friction necessary for the car to hold this last radius of curvature without skidding? Picture the Problem (a) We can apply Newton’s second law to the car to find the force of static friction exerted on the car by the surface and use the definition of the coefficient of static friction to find the minimum coefficient of static friction necessary for the car to sustain its speed. In Part (b), we can again apply Newton’s second law, this time in both tangential and radial form, to find the values of r for the given angles.


477

(a) The forces acting on the car as it moves at a constant speed of 100 km/h are shown in the pictorial representation to the right. x

y

nFr

dFr

gmr

sfr

Apply aFrr

m=∑ to the car to obtain:

0ds =−=∑ FfFx (1)

and 0n =−=∑ mgFFy (2)

Solving equation (1) for fs yields: N 500ds == Ff

Use the definition of the coefficient of static friction to obtain:

n

smin s, F

f=μ

From equation (2), mgF =n ; hence: mg

fsmin s, =μ

Substitute numerical values and evaluate min s,μ :

( )( )

0255.0

m/s 81.9kg 2000N 500

2min s,

=

=μ

(b) The horizontal forces acting on the car (shown as a top view) as it travels clockwise along a path of radius of curvature r are shown in the diagram to the right.

dFr

sfr

θ

r

Apply cradial maF =∑ to the car to

obtain:

rvmf

2

s sin =θ (1)

Chapter 5

478

Appling ttangential maF =∑ to the car

yields:

0cos ds =− Ff θ or

ds cos Ff =θ (2)

Divide equation (1) by equation (2):

d

2

tanrFmv

=θ ⇒θtand

2

Fmvr =

Substitute numerical values and evaluate r for θ = 45.0°:

( )

( )km .093

0.45tanN 500s 3600

h 1km

m 10h

km100kg 200023

=

°

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅

=r

Substitute numerical values and valuate r for θ = 88.0°:

( )

( )m 108m 8.107

0.88tanN 500s 3600

h 1km

m 10h

km100kg 200023

==

°

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅

=r

(c) The minimum coefficient of static friction necessary for the car to hold this last radius of curvature without skidding is given by:

rgv

mgr

vm

mgf 2

2

min s,min s, ===μ

Substitute numerical values and evaluate min s,μ :

( )( )730.0

m/s 81.9m 07.81s 3600

h 1km

m 10h

km100

2

23

min s,

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅

=μ

91 •• Suppose you ride a bicycle in a 20-m-radius circle on a horizontal surface. The resultant force exerted by the surface on the bicycle (normal force plus frictional force) makes an angle of 15º with the vertical. (a) What is your speed? (b) If the frictional force on the bicycle is half its maximum possible value, what is the coefficient of static friction?


479

Picture the Problem The forces acting on the bicycle are shown in the force diagram. The static friction force is the centripetal force exerted by the surface on the bicycle that allows it to move in a circular path. sn fF

rr+ makes an angle

θ with the vertical direction. The application of Newton’s second law will allow us to relate this angle to the speed of the bicycle and the coefficient of static friction.

x

y

nFr

gmFrr

=g

sfr

θ

(a) Apply ∑ = aF

rrm to the bicycle:

rmvfFx

2

s ==∑

and 0n =−=∑ mgFFy

Relate Fn and fs to θ :

rgv

mgr

mv

Ff 2

2

n

stan ===θ

Solving for v yields: θtanrgv =


( )( )m/s 7.3

tan15m/s9.81m20 2

=

°=v

(b) Relate fs to μs and Fn: mgff s2

1maxs,2

1s μ==

Solve for μs and substitute for fs to obtain:

rgv

mgf 2

ss

22==μ

Substitute numerical values and evaluate μs

( )( )( ) 54.0

m/s9.81m20m/s7.252

2

2

s ==μ

92 •• An airplane is flying in a horizontal circle at a speed of 480 km/h. The plane is banked for this turn, its wings tilted at an angle of 40º from the horizontal (Figure 5-76). Assume that a lift force acting perpendicular to the wings acts on the aircraft as it moves through the air. What is the radius of the circle in which the plane is flying?

Chapter 5

480

Picture the Problem The diagram shows the forces acting on the plane as it flies in a horizontal circle of radius r. We can apply Newton’s second law to the plane and eliminate the lift force in order to obtain an expression for r as a function of v and θ.

liftFr

gmFrr

=gθ

θ

x

y

r

Apply aF

rrm=∑ to the plane:

rvmFFx

2

lift sin ==∑ θ

and 0coslift =−=∑ mgFFy θ

Eliminate Flift between these equations to obtain:

rgv2

tan =θ ⇒θtan

2

gvr =

Substitute numerical values and evaluate r: ( ) km 2.2

tan40m/s9.81s3600

h1h

km480

2

2

=°

⎟⎟⎠

⎞⎜⎜⎝

⎛×

=r

93 •• An automobile club plans to race a 750-kg car at the local racetrack. The car needs to be able to travel around several 160-m-radius curves at 90 km/h. What should the banking angle of the curves be so that the force of the pavement on the tires of the car is in the normal direction? (Hint: What does this requirement tell you about the frictional force?) Picture the Problem Under the conditions described in the problem statement, the only forces acting on the car are the normal force exerted by the road and the gravitational force exerted by Earth. The horizontal component of the normal force is the centripetal force. The application of Newton’s second law will allow us to express θ in terms of v, r, and g.

nFr

x

y

θ

θ

gmFrr

=g


481

Apply ∑ = aFrr

m to the car:

∑

∑

=−=

==

0cosand

sin

n

2

n

mgFF

rvmFF

y

x

θ

θ

Eliminate Fn from the force equations to obtain: rg

v2

tan =θ ⇒ ⎥⎦

⎤⎢⎣

⎡= −

rgv2

1tanθ

Substitute numerical values and evaluate θ: ( )( ) °=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧⎟⎠⎞

⎜⎝⎛ ×

= − 22m/s9.81m160

s 3600h 1

hkm 90

tan 2

2

1θ

94 •• A curve of radius 150 m is banked at an angle of 10º. An 800-kg car negotiates the curve at 85 km/h without skidding. Neglect the effects of air drag and rolling friction. Find (a) the normal force exerted by the pavement on the tires, (b) the frictional force exerted by the pavement on the tires, (c) the minimum coefficient of static friction between the pavement and the tires. Picture the Problem Both the normal force and the static friction force contribute to the centripetal force in the situation described in this problem. We can apply Newton’s second law to relate fs and Fn and then solve these equations simultaneously to determine each of these quantities.

nFr

x

y

θ

θ

gmFrr

=g

θ

sfr

(a) Apply ∑ = aF

rrm to the car: ∑ =+=

rvmfFFx

2

sn cossin θθ

and ∑ =−−= 0sincos sn mgfFFy θθ

Multiply the x equation by sinθ and the y equation by cosθ to obtain: θθθθ sinsincossin

22

ns rvmFf =+

and 0coscossincos s

2n =−− θθθθ mgfF

Chapter 5

482

Add these equations to eliminate fs: θθ sincos2

n rvmmgF =−

Solve for Fn:

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

+=

θθ

θθ

sincos

sincos

2

2

n

rvgm

rvmmgF


( ) ( )

kN 8.3

kN 245.8sin10m150

s 3600h 1

hkm 85

cos10m/s9.81kg800

2

2n

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

°⎟⎠⎞

⎜⎝⎛ ×

+°=F

(b) Solve the y-equation for fs:

θθ

sincosn

smgFf −

=


( ) ( )( ) kN 6.1kN565.1sin10

m/s9.81kg800cos10kN8.245 2

s ==°

−°=f

(c) Express μs,min in terms of fs and Fn: n

smins, F

f=μ

Substitute numerical values and evaluate μs,min:

19.0kN8.245kN1.565

mins, ==μ

95 •• On another occasion, the car in Problem 94 negotiates the curve at 38 km/h. Neglect the effects of air drag and rolling friction. Find (a) the normal force exerted on the tires by the pavement, and (b) the frictional force exerted on the tires by the pavement.


483

Picture the Problem Both the normal force and the static friction force contribute to the centripetal force in the situation described in this problem. We can apply Newton’s second law to relate fs and Fn and then solve these equations simultaneously to determine each of these quantities.

nFr

x

y

θ

θ

gmFrr

=g

θ

sfr

(a) Apply ∑ = aF

rrm to the car: ∑ =+=

rvmfFFx

2

sn cossin θθ

and ∑ =−−= 0sincos sn mgfFFy θθ

Multiply the x equation by sinθ and the y equation by cosθ : θθθθ sinsincossin

22

ns rvmFf =+

0coscossincos s2

n =−− θθθθ mgfF

Add these equations to eliminate fs: θθ sincos2

n rvmmgF =−

Solving for Fn yields:

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

+=

θθ

θθ

sincos

sincos

2

2

n

rvgm

rvmmgF


( ) ( )

kN 8.7

kN 83.7sin10m150

s 3600h 1

hkm 38

cos10m/s9.81kg800

2

2n

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

°⎟⎠⎞

⎜⎝⎛ ×

+°=F

(b) Solve the y equation for fs:

θθ

θθ

sincot

sincos

nn

smgFmgFf −=

−=

Chapter 5

484


( ) ( )( ) kN 78.0sin10

m/s9.81kg80010cotkN 83.72

s −=°

−°=f

The negative sign tells us that fs points upward along the inclined plane rather than as shown in the force diagram. 96 ••• As a civil engineer intern during one of your summers in college, you are asked to design a curved section of roadway that meets the following conditions: When ice is on the road, and the coefficient of static friction between the road and rubber is 0.080, a car at rest must not slide into the ditch and a car traveling less than 60 km/h must not skid to the outside of the curve. Neglect the effects of air drag and rolling friction. What is the minimum radius of curvature of the curve and at what angle should the road be banked? Picture the Problem The free-body diagram to the left is for the car at rest. The static friction force up the incline balances the downward component of the car’s weight and prevents it from sliding. In the free-body diagram to the right, the static friction force points in the opposite direction as the tendency of the moving car is to slide toward the outside of the curve.

nFr

x

y

θ

θ

gmFrr

=g

θ

nFr

x

y

θ

θ

gmFrr

=g

θmaxs,fr

maxs,fr

0=v 0>v

Apply ∑ = aF

rrm to the car that is at

rest: ∑ =−+= 0sincos sny mgfFF θθ (1)

and ∑ =−= 0cossin snx θθ fFF (2)

Substitute fs = fs,max = μsFn in equation (2) and solve for the maximum allowable value of θ :

( )s1tan μθ −=


( ) °=°== − 6.457.4080.0tan 1θ


485

Apply∑ = aFrr

m to the car that is

moving with speed v: ∑ =−−= 0sincos sny mgfFF θθ (3)

∑ =+=rv

mfFF2

snx cossin θθ (4)

Substitute fs = μsFn in equations (3) and (4) and simplify to obtain:

( ) mgF =− θμθ sincos sn (5) and

( )rvmcF

2

sn sinos =+ θθμ (6)

Substitute numerical values in equations (5) and (6) to obtain:

0.9904Fn = mg and

rvmF

2

n1595.0 =

Eliminate Fn between these equations and solve for r: g

vr1610.0

2

=

Substitute numerical values and evaluate r:

( ) km 18.0m/s9.810.1610

s 3600h 1

hkm 60

2

2

=⎟⎠⎞

⎜⎝⎛ ×

=r

97 ••• A curve of radius 30 m is banked so that a 950-kg car traveling at 40.0 km/h can round it even if the road is so icy that the coefficient of static friction is approximately zero. You are commissioned to tell the local police the range of speeds at which a car can travel around this curve without skidding. Neglect the effects of air drag and rolling friction. If the coefficient of static friction between the road and the tires is 0.300, what is the range of speeds you tell them? Picture the Problem The free-body diagram to the left is for the car rounding the curve at the minimum (not sliding down the incline) speed. The combination of the horizontal components of the frictional force and the normal force provides the net force toward the center of the curve. In the free-body diagram to the right, the static friction force points in the opposite direction as the tendency of the car moving with the maximum safe speed is to slide toward the outside of the curve. Application of Newton’s second law and the simultaneous solution of the force equations will yield vmin and vmax.

Chapter 5

486

nFr

x

y

θ

θ

gmFrr

=g

θsfr

maxvv =nFr

x

y

θ

θ

gmFrr

=g

sfr

θminvv =

Apply ∑ = aF

rrm to a car traveling

around the curve when the coefficient of static friction is zero:

rvmFF

2min

nx sin∑ == θ

and 0cosny∑ =−= mgFF θ

Divide the first of these equations by the second to obtain: rg

v2

tan =θ ⇒ ⎟⎟⎠

⎞⎜⎜⎝

⎛= −

rgv2

1tanθ

Substitute numerical values and evaluate the banking angle:

( )( ) °=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛××

= − 76.22m/s81.9m30

s 3600h 1

kmm10

hkm40

tan 2

23

1θ

Apply∑ = aF

rrm to the car traveling

around the curve at minimum speed: rvmfFF

2min

snx cossin∑ =−= θθ

and 0sincos sny∑ =−+= mgfFF θθ

Substitute nsmaxs,s Fff μ== in

the force equations and simplify to obtain:

( )r

vmF2min

sn cossin =− θμθ

and ( ) mgF =+ θμθ sincos sn

Evaluate these equations for θ = 22.876 and μs = 0.300: 0.1102Fn= r

vm2min and 1.038Fn = mg


487

Eliminate Fn between these two equations and solve for minv :

rgv 106.0min =

Substitute numerical values and evaluate minv :

( )( )km/h 20m/s 6.5

m/s9.81m30106.0 2min

≈=

=v

Apply∑ = aF

rrm to the car traveling

around the curve at maximum speed: rvmfFF

2max

snx cossin∑ =+= θθ

and 0sincos sny∑ =−−= mgfFF θθ

Substitute nsmax s,s Fff μ== in the

force equations and simplify to obtain:

( )r

vmF2max

sn sincos =+ θθμ

and ( ) mgF =− θμθ sincos sn

Evaluate these equations for θ = 22.76° and μs = 0.300: r

vmF2max

n6635.0 =

and mgF 8061.08061.0 n =

Eliminate Fn between these two equations and solve for vmax:

rgv 8231.0max =

Substitute numerical values and evaluate vmax:

( )( )( )km/h 65m/s 61

m/s81.9m308231.0 2max

≈=

=v

You should tell them that the safe range of speeds is km/h 56 km/m 20 ≤≤ v . Euler’s Method 98 •• You are riding in a hot air balloon when you throw a baseball straight down with an initial speed of 35.0 km/h. The baseball falls with a terminal speed of 150 km/h. Assuming air drag is proportional to the speed squared, use Euler’s method (spreadsheet) to estimate the speed of the ball after 10.0 s. What is the uncertainty in this estimate? You drop a second baseball, this one released from rest. How long does it take for it to reach 99 percent of its terminal speed? How far does it fall during this time?

Chapter 5

488

Picture the Problem The free-body diagram shows the forces acting on the baseball sometime after it has been thrown downward but before it has reached its terminal speed. In order to use Euler’s method, we’ll need to determine how the acceleration of the ball varies with its speed. We can do this by applying Newton’s second law to the ball and using its terminal speed to express the constant in the acceleration equation in terms of the ball’s terminal speed. We can then use

tavv nnn Δ+=+1 to find the speed of the ball at any given time.

jcvF ˆ2d −=r

jmgF ˆg =r

y

Apply Newton’s second law to the ball to obtain: dt

dvmbvmg =− 2 ⇒ 2vmbg

dtdv

−= (1)

When the ball reaches its terminal speed, its acceleration is zero:

2t0 v

mbg −= ⇒ 2

tvg

mb=

Substitute in equation (1) to obtain: ⎟⎟

⎠

⎞⎜⎜⎝

⎛−= 2

t

2

1vvg

dtdv

Express the position of the ball to obtain:

tvvxx nnnn Δ

++= +

+ 21

1

Letting an be the acceleration of the ball at time tn, express its speed when t = tn + 1:

tavv nnn Δ+=+1

where ⎟⎟⎠

⎞⎜⎜⎝

⎛−= 2

t

2

1vvga n

n and Δt is an

arbitrarily small interval of time.

A spreadsheet solution is shown below. The formulas used to calculate the quantities in the columns are as follows:

Cell Formula/Content Algebraic Form A10 B9+$B$1 t + Δt B10 B9+0.5*(C9+C10)*$B$1

tvvxx nnnn Δ

++= +

+ 21

1

C10 C9+D9*$B$1 vn+1 = vn+ anΔt D10 $B$4*(1−C10^2/$B$5^2)

⎟⎟⎠

⎞⎜⎜⎝

⎛−= 2

t

2

1vvga n

n


489

A B C D 1 Δt= 0.5 s 2 x0= 0 m 3 v0= 9.722 m/s 4 a0= 9.81 m/s2 5 vt= 41.67 m/s 6 7 t x v a 8 (s) (m) (m/s) (m/s2) 9 0.0 0 9.7 9.28 10 0.5 6 14.4 8.64 11 1.0 14 18.7 7.84 12 1.5 25 22.6 6.92

28 9.5 317 41.3 0.17 29 10.0 337 41.4 0.13 30 10.5 358 41.5 0.10

38 14.5 524 41.6 0.01 39 15.0 545 41.7 0.01 40 15.5 566 41.7 0.01 41 16.0 587 41.7 0.01 42 16.5 608 41.7 0.00

From the table we can see that the speed of the ball after 10 s is approximately

m/s.4.41 We can estimate the uncertainty in this result by halving Δt and

recalculating the speed of the ball at t = 10 s. Doing so yields v(10 s) ≈ 41.3 m/s, a difference of about %.02.0

Chapter 5

490

The following graph shows the velocity of the ball thrown straight down as a function of time.

Ball Thrown Straight Down

0

5

10

15

20

25

30

35

40

45

0 5 10 15 20

t (s)

v (m

/s)

Reset Δt to 0.5 s and set v0 = 0. Ninety-nine percent of 41.67 m/s is approximately 41.3 m/s. Note that the ball will reach this speed in about s5.10 and that the

distance it travels in this time is about m.322 The following graph shows the distance traveled by the ball dropped from rest as a function of time.

Ball Dropped From Rest

0

50

100

150

200

250

300

350

400

0 2 4 6 8 10 12

t (s)

x(m

)

99 •• [SSM] You throw a baseball straight up with an initial speed of 150 km/h. The ball’s terminal speed when falling is also 150 km/h. (a) Use Euler’s method (spreadsheet) to estimate its height 3.50 s after release. (b) What is the maximum height it reaches? (c) How long after release does it reach its maximum height? (d) How much later does it return to the ground? (e) Is the time the ball spends on the way up less than, the same as, or greater than the time it spends on the way down?


491

Picture the Problem The free-body diagram shows the forces acting on the baseball after it has left your hand. In order to use Euler’s method, we’ll need to determine how the acceleration of the ball varies with its speed. We can do this by applying Newton’s second law to the baseball. We can then use

tavv nnn Δ+=+1 and tvxx nnn Δ+=+1 to find the speed and position of the ball.

jbvF ˆ2d −=r

y

jmgF ˆg −=r

Apply ∑ = yy maF to the baseball:

dtdvmmgvbv =−−

where vv = for the upward part of the

flight of the ball and vv −= for the downward part of the flight.

Solve for dv/dt to obtain: vv

mbg

dtdv

−−=

Under terminal speed conditions ( tvv −= ):

2t0 v

mbg +−= and 2

tvg

mb=

Substituting for b/m yields:

⎟⎟⎠

⎞⎜⎜⎝

⎛+−=−−= 2

t2t

1vvv

gvvvgg

dtdv

Letting an be the acceleration of the ball at time tn, express its position and speed when t = tn + 1:

( ) tvvyy nnnn Δ++= −+ 121

1 and

tavv nnn Δ+=+1

where ⎟⎟⎠

⎞⎜⎜⎝

⎛+−= 2

t

1vvv

ga nnn and Δt is an



Cell Formula/Content Algebraic Form D11 D10+$B$6 t + Δt E10 41.7 v0 E11 E10−$B$4* tavv nnn Δ+=+1

Chapter 5

492

(1+E10*ABS(E10)/($B$5^2))*$B$6F10 0 y0 F11 F10+0.5*(E10+E11)*$B$6 ( ) tvvyy nnnn Δ++= −+ 12

11

G10 0 y0 G11 $E$10*D11−0.5*$B$4*D11^2 2

21

0 gttv −

A B C D E F G 4 g= 9.81 m/s2 5 vt= 41.7 m/s 6 Δt= 0.1 s 7 8 9 t v y yno drag 10 0.00 41.70 0.00 0.00 11 0.10 39.74 4.07 4.12 12 0.20 37.87 7.95 8.14

40 3.00 3.01 60.13 81.00 41 3.10 2.03 60.39 82.18 42 3.20 1.05 60.54 83.26 43 3.30 0.07 60.60 84.25 44 3.40 −0.91 60.55 85.14 45 3.50 −1.89 60.41 85.93 46 3.60 −2.87 60.17 86.62

78 6.80 −28.34 6.26 56.98 79 6.90 −28.86 3.41 54.44 80 7.00 −29.37 0.49 51.80 81 7.10 −29.87 −2.47 49.06

(a) When t = 3.50 s, the height of the ball is about m4.60 . (b) The maximum height reached by the ball is m6.60 .

(c) The time the ball takes to reach its maximum height is about s 0.3 .

(d) The ball hits the ground at about t = s0.7

(e) Because the time the ball takes to reach its maximum height is less than half its time of flight, the time the ball spends on the way up less than the time it spends on the way down


493

100 ••• A 0.80-kg block on a horizontal frictionless surface is held against a massless spring, compressing it 30 cm. The force constant of the spring is 50 N/m. The block is released and the spring pushes it 30 cm. Use Euler’s method (spreadsheet) with Δt = 0.005 s to estimate the time it takes for the spring to push the block the 30 cm. How fast is the block moving at this time? What is the uncertainty in this speed? Picture the Problem The pictorial representation shows the block in its initial position against the compressed spring, later as the spring accelerates it to the right, and finally when it has reached its maximum speed at xf = 0.30 m. In order to use Euler’s method, we’ll need to determine how the acceleration of the block varies with its position. We can do this by applying Newton’s second law to the block. We can then use tavv nnn Δ+=+1 and tvxx nnn Δ+=+1 to find the speed and position of the block.

Apply ∑ = xx maF to the block:

( ) nn maxk =−m30.0

Solve for an: ( )nn xmka −= m30.0

Express the position and speed of the block when t = tn + 1:

tvxx nnn Δ+=+1 and

tavv nnn Δ+=+1

where ( )nn xmka −= m30.0 and Δt is an



Cell Formula/Content Algebraic FormA10 A9+$B$1 t + Δt B10 B9+C10*$B$1 tvx nn Δ+ C10 C9+D9*$B$1 tav nn Δ+ D10 ($B$4/$B$5)*(0.30−B10) ( )nx

mk

−30.0

Chapter 5

494

A B C D 1 Δt= 0.005 s 2 x0= 0 m 3 v0= 0 m/s 4 k = 50 N/m 5 m = 0.80 kg 6 7 t x v a 8 (s) (m) (m/s) (m/s2) 9 0.000 0.00 0.00 18.75 10 0.005 0.00 0.09 18.72 11 0.010 0.00 0.19 18.69 12 0.015 0.00 0.28 18.63

45 0.180 0.25 2.41 2.85 46 0.185 0.27 2.42 2.10 47 0.190 0.28 2.43 1.34 48 0.195 0.29 2.44 0.58 49 0.200 0.30 2.44 −0.19

From the table we can see that it took about s200.0 for the spring to push the

block 30 cm and that it was traveling about m/s44.2 at that time. We can

estimate the uncertainty in this result by halving Δt and recalculating the speed of the ball at t = 10 s. Doing so yields v(0.20 s) ≈ 2.41 m/s, a difference of about

%.2.1 Finding the Center of Mass 101 • Three point masses of 2.0 kg each are located on the x axis. One is at the origin, another at x = 0.20 m, and another at x = 0.50 m. Find the center of mass of the system. Picture the Problem We can use its definition to find the center of mass of this system. The x coordinate of the center of mass is given by:

321

332211cm mmm

xmxmxmx++++

=

Substitute numerical values and evaluate xcm:

( )( ) ( )( ) ( )( ) m23.0kg2.0kg2.0kg2.0

m0.50kg2.0m0.20kg2.00kg2.0cm =

++++

=x


495

Because the point masses all lie along the x axis:

0cm =y and the center of mass of this

system of particles is at ( )0,m23.0 .

102 • On a weekend archeological dig, you discover an old club-ax that consists of a symmetrical 8.0-kg stone attached to the end of a uniform 2.5-kg stick. You measure the dimensions of the club-ax as shown in Figure 5-77. How far is the center of mass of the club-ax from the handle end of the club-ax? Picture the Problem Let the left end of the handle be the origin of our coordinate system. We can disassemble the club-ax, find the center of mass of each piece, and then use these coordinates and the masses of the handle and stone to find the center of mass of the club-ax. Express the center of mass of the handle plus stone system:

stonestick

stonecm,stonestickcm,stickcm mm

xmxmx

+

+=

Assume that the stone is drilled and the stick passes through it. Use symmetry considerations to locate the center of mass of the stick:

cm49stickcm, =x

Use symmetry considerations to locate the center of mass of the stone:

cm89stonecm, =x


( )( ) ( )( )

cm 79

kg8.0kg2.5cm89kg8.0cm94kg2.5

cm

=

++

=x

103 • Three balls A, B, and C, with masses of 3.0 kg, 1.0 kg, and 1.0 kg, respectively, are connected by massless rods, as shown in Figure 5-78. What are the coordinates of the center of mass of this system? Picture the Problem We can treat each of balls as though they are point objects and apply the definition of the center of mass to find (xcm, ycm). The x coordinate of the center of mass is given by: CBA

CCBBAAcm mmm

xmxmxmx++++

=

Chapter 5

496


( )( ) ( )( ) ( )( ) m 0.2kg1.0kg1.0kg3.0

m3.0kg1.0m1.0kg1.0m2.0kg3.0cm =

++++

=x

The y coordinate of the center of mass is given by: CBA

CCBBAAcm mmm

ymymymy++++

=

Substitute numerical values and evaluate ycm:

( )( ) ( )( ) ( )( ) m 4.1kg1.0kg1.0kg3.0

0kg1.0m1.0kg1.0m2.0kg3.0cm =

++++

=y

The center of mass of this system of balls is at ( )m4.1,m0.2 .

Alternate Solution Using Vectors Picture the Problem We can use the vector expression for the center of mass to find (xcm, ycm). The vector expression for the center of mass is: iii

mM rrrr Σcm = or

M

m iiir

r

rr Σcm = (1)

where kjir ˆˆˆ

cmcmcmcm zyx ++=r

The position vectors for the objects located at A, B, and C are:

( )m ˆ0ˆ0.2ˆ0.2A kjir ++=r

, kjir ˆ0ˆˆ

B ++=r , and

( )m ˆ0ˆ0ˆ0.3C kjir ++=r

Substitute numerical values in (1) and simplify to obtain:

( ) ( )( ) ( )( )[( )( ) ]

( ) ( ) kji

kji

kjikjir

ˆ0ˆm 4.1ˆm 0.2

m ˆ0ˆ0ˆ0.3kg 0.1

m ˆ0ˆˆkg 0.1mˆ0ˆ0.2ˆ0.2kg 0.3kg 1.0 kg 1.0kg0.3

1cm

++=

+++

+++++++

=r

The center of mass of this system of balls is at ( )0 m,4.1,m0.2 .


497

104 • By symmetry, locate the center of mass of an equilateral triangle with edges of length a. The triangle has one vertex on the y axis and the others at (–a/2, 0) and (+a/2, 0). Picture the Problem The figure shows an equilateral triangle with its y-axis vertex above the x axis. The bisectors of the vertex angles are also shown. We can find x coordinate of the center-of-mass by inspection and the y coordinate using trigonometry. From symmetry considerations:

0cm =x

Express the trigonometric relationship between a/2, 30°, and

cmy :

ay

21

cm30tan =°

Solve for cmy and simplify to obtain: aay 29.030tan21

cm =°=

The center of mass of an equilateral triangle oriented as shown above is at ( )a29.0,0 .

105 •• [SSM] Find the center of mass of the uniform sheet of plywood in Figure 5-79. Consider this as a system of effectively two sheets, letting one have a ″negative mass″ to account for the cutout. Thus, one is a square sheet of 3.0-m edge length and mass m1 and the second is a rectangular sheet measuring 1.0 m × 2.0 m with a mass of –m2. Let the coordinate origin be at the lower left corner of the sheet. Picture the Problem Let the subscript 1 refer to the 3.0-m by 3.0-m sheet of plywood before the 2.0-m by 1.0-m piece has been cut from it. Let the subscript 2 refer to 2.0-m by 1.0-m piece that has been removed and let σ be the area density of the sheet. We can find the center-of-mass of these two regions; treating the missing region as though it had negative mass, and then finding the center-of-mass of the U-shaped region by applying its definition.

Chapter 5

498

Express the coordinates of the center of mass of the sheet of plywood: 21

2,cm21cm,1cm mm

xmxmx

−−

=

and

21

2,cm21cm,1cm mm

ymymy

−−

=

Use symmetry to find xcm,1, ycm,1, xcm,2, and ycm,2:

m0.2,m5.1and

m5.1m,5.1

cm,2cm,2

cm,1cm,1

==

==

yx

yx

Determine m1 and m2:

kgAm

Am

σσ

σσ

2and

kg9

22

11

==

==

Substitute numerical values and evaluate cmx :

( )( ) ( )( )

m5.1kg2kg9

kg5.1kg2m5.1kg9cm

=−−

=σσσσx

Substitute numerical values and evaluate cmy :

( )( ) ( )( )

m4.1kg2kg9

m0.2kg2m5.1kg9cm

=−−

=σσσσy

The center of mass of the U-shaped sheet of plywood is at ( )m1.4m,1.5 .

106 •• A can in the shape of a symmetrical cylinder with mass M and height H is filled with water. The initial mass of the water is M, the same mass as the can. A small hole is punched in the bottom of the can, and the water drains out. (a) If the height of the water in the can is x, what is the height of the center of mass of the can plus the water remaining in the can? (b) What is the minimum height of the center of mass as the water drains out? Picture the Problem We can use its definition to find the center of mass of the can plus water. By setting the derivative of this function equal to zero, we can find the value of x that corresponds to the minimum height of the center of mass of the water as it drains out and then use this extreme value to express the minimum height of the center of mass. (a) Using its definition, express the location of the center of mass of the can + water: mM

xmHMx

+

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

= 22cm (1)


499

Let the cross-sectional area of the can be A and use the definition of density to relate the mass m of water remaining in the can at any given time to its depth x:

Axm

AHM

==ρ ⇒ MHxm =

Substitute for m in equation (1) and simplify to obtain:

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

+

⎟⎠⎞

⎜⎝⎛+

=+

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

=

Hx

Hx

H

MHxM

xMHxHM

x1

1

222

2

cm

(b) Differentiate cmx with respect to x and set the derivative equal to zero for extrema:

021

12

1

21

121

21

21

2cm =

+

+

−

+

+=

+

+=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

⎪⎪

⎩

⎪⎪

⎨

⎧

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛⎟⎠⎞

⎜⎝⎛

Hx

HHx

Hx

HHx

Hx

H

Hx

Hx

dxdH

dxdx

Simplify this expression to obtain a quadratic equation in x/H: 012

2

=−⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

Hx

Hx

Solving for x/H yields:

( ) HHx 414.012 ≈−= where we’ve kept the positive solution because a negative value for x/H would make no sense.

Use your graphing calculator to convince yourself that the graph of cmx as a function of x is concave upward at Hx 414.0≈ and that, therefore, the minimum value of cmx occurs at .414.0 Hx ≈ Evaluate cmx at ( )12 −= Hx to obtain:

( )

( )

( )

( )12

121

121

2

2

12cm

−=

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−+

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+

=−=

H

HH

HH

HxHx

Chapter 5

500

107 •• [SSM] Two identical thin uniform rods each of length L are glued together so that the angle at the joint is 90°. Determine the location of the center of mass (in terms of L) of this configuration relative to the origin taken to be at the joint. (Hint: You do not need the mass of the rods (why?), but you should start by assuming a mass m and see that it cancels out.) Picture the Problem A pictorial representation of the system is shown to the right. The x and y coordinates of the two rods are

( ) ( )Lyx 21

cm1,cm1, ,0, = and

( ) ( ). 0,, 21

cm2,cm2, Lyx = We can use the definition of the center of mass to find the coordinates ( )., cmcm yx x

y

L

LLx 21

cm2, =

Ly 21

cm1, =

( )cmcm, yx

2m

1m

The x coordinate of the center of mass is given by: 21

2cm2,1cm1,cm mm

mxmxx

++

=


( ) ( )21

221

1cm

0mm

mLmx

++

=

or, because m1 = m2 = m, ( ) ( )

Lmm

mLmx 4

121

cm0

=++

=

The y coordinate of the center of mass is given by: 21

2cm2,1cm1,cm mm

mymyy

++

=


( ) ( )21

2121

cm0

mmmmL

y++

=

or, because m1 = m2 = m, ( ) ( )

Lmm

mmLy 4

121

cm0

=++

=

The center of mass of this system is located at ( )LL 4

141 , .

Remarks: Note that the center of mass is located at a distance °= 45cos2

1 Ld from the vertex on the axis of symmetry bisecting the two arms.


501

108 ••• Repeat the analysis of Problem 107 with a general angle θ at the joint instead of 90°. Does your answer agree with the specific 90°- angle answer in Problem 107 if you set θ equal to 90°? Does your answer give plausible results for angles of zero and 180°? Picture the Problem The pictorial representation of the system is shown to the right. The x and y coordinates of the two rods are

( ) ( )θθ sin ,cos, 21

21

cm1,cm1, LLyx = and

( ) ( ). 0,, 21

cm2,cm2, Lyx = We can use the definition of the center of mass to find the coordinates ( )., cmcm yx

x

y

L

Lx cm2,

( )cmcm, yxθ

cosL θ

sinL θ

cm1,x

cm1,y

2m

1m

The x coordinate of the center of mass is given by: 21

2cm2,1cm1,cm mm

mxmxx

++

=


( ) ( )21

221

121

cmcos

mmmLmL

x++

=θ

or, because m1 = m2 = m, ( ) ( )

( )θ

θ

cos1

cos

41

21

21

cm

+=++

=

Lmm

mLmLx

The y coordinate of the center of mass is given by: 21

2cm2,1cm1,cm mm

mymyy

++

=


( ) ( )21

2121

cm0sin

mmmmL

y+

+=

θ

or, because m1 = m2 = m, ( ) ( )

θθ

sin0sin

412

1

cm Lmm

mmLy =

++

=

The center of mass of this system of rods is located at ( )( )θθ sin ,cos1 4

141 LL + .

For °= 0θ :

( )( ) ( )( ) ( ) expected. as 0,0sin,0cos1sin ,cos1 21

41

41

41

41 LLLLL =°°+=+ θθ

Chapter 5

502

For °= 90θ :

( )( ) ( )( )( ) 107. Problemwith agreement in ,

90sin,90cos1sin ,cos1

41

41

41

41

41

41

LL

LLLL

=

°°+=+ θθ

For °= 180θ :

( )( ) ( )( ) ( ) expected. as 0,0180sin,180cos1sin ,cos1 41

41

41

41 =°°+=+ LLLL θθ

Remarks: Note that the center of mass is located at a distance

θcos1241 += Ld from the vertex on an axis that makes an angle

⎥⎦⎤

⎢⎣⎡+

= −

θθφ

cos1sintan 1 with the x axis.

*Finding the Center of Mass by Integration 109 •• Show that the center of mass of a uniform semicircular disk of radius R is at a point 4R/(3π) from the center of the circle. Picture the Problem A semicircular disk and a surface element of area dA is shown in the diagram. Because the disk is a continuous object, we’ll use

∫= dmM rrrr

cm and symmetry to find its

center of mass. R

y

xθ

dA

Express the coordinates of the center of mass of the semicircular disk:

symmetry.by0cm =x

M

dAyy ∫=

σcm (1)

Express y as a function of r and θ : θsinry =

Express dA in terms of r and θ :

( )( )θθ

θθθθ

dRdRR

dyrdA

22 cos2coscos2

cos2

=

==


503

Express M as a function of r and θ : 2

21

diskhalf RAM σπσ == ⇒ 22

RM

πσ =

Substitute for y, σ and dA in equation (1) to obtain: ( )( )∫=

2

0

222cm cos2sin2π

θθθπ

dRRR

y

Simplifying yields: ∫=

2

0

2cm cossin4

π

θθθπ

dRy

Change variables by letting

θcos=u . Then:

θθ ddu sin−= and

2

1

2

13

44 32

cm

l

l

l

l

⎥⎦

⎤⎢⎣

⎡−=−= ∫

uRduuRyππ

Substituting for u gives:

π

πθ

π

π

34

314

3cos4 2

0

3

cm

R

RRy

=

⎥⎦⎤

⎢⎣⎡−−=⎥

⎦

⎤⎢⎣

⎡−=

110 •• Find the location of the center of mass of a nonuniform rod 0.40 m in length if its density varies linearly from 1.00 g/cm at one end to 5.00 g/cm at the other end. Specify the center-of-mass location relative to the less-massive end of the rod. Picture the Problem The pictorial representation summarizes the information we’re given about the non-uniform rod. We can use the definition of the center of mass for a continuous object to find the center of mass of the non-uniform rod.

40

x, cm

0μ(x) = 1.00 g/cm2 (x) = 5.00 g/cmμ 2

dm

dx

Chapter 5

504

The x coordinate of the center of mass of the non-uniform rod is given by:

∫∫=

dm

xdmxcm

or, because ( ) ,dxxdm μ= ( )( )∫

∫=dxx

dxxxx

μ

μcm (1)

By symmetry:

0cm =y

Use the given information regarding the linear variation in the density of the non-uniform rod to express μ(x):

( ) ( )xx 2g/cm 0.10g/cm 00.1 +=μ

Substituting for μ(x) in equation (1) and evaluating the integrals yields:

( )[ ]

( )[ ]cm 24

g/cm 0.10g/cm 00.1

g/cm 0.10g/cm 00.1

cm 40

0

2

cm 40

0

2

cm =+

+=

∫

∫

dxx

xdxxxx

The coordinates of the center of mass of the non-uniform rod are ( )0 cm, 24 . 111 ••• You have a thin uniform wire bent into part of a circle that is described by a radius R and angle θm (see Figure 5-80). Show that the location of its center of mass is on the x axis and located a distance m mcm ( sin )/x R θ θ= , where θm is expressed in radians. Double check by showing that this answer gives the physically expected limit for θm = 180°. Verify that your answer gives you the result in the text (in the subsection Finding the Center of Mass by Integration) for the special case of θ m = 90°.


505

Picture the Problem We can use the definition of the center of mass for a continuous object to find the center of mass of the non-uniform rod.

θ

θ

θ

θm

m

x

y

Rdsdmd

The x coordinate of the center of mass of the thin uniform wire is given by:

∫∫=

dm

xdmxcm

or, because ,θλλ Rddsdm ==

∫

∫

∫

∫

−

−

−

− ==m

m

m

m

m

m

m

mcm θ

θ

θ

θθ

θ

θ

θ

θ

θ

θλ

θλ

d

xd

Rd

Rdxx (1)

By symmetry:

0cm =y

Because θcosRx = , equation (1) becomes:

∫

∫

∫

∫

−

−

−

− ==m

m

m

m

m

m

m

m

coscos

cm θ

θ

θ

θθ

θ

θ

θ

θ

θθ

θ

θθ

d

dR

d

dRx

Evaluate these integrals to obtain:

m

mcm

sinθθRx =

The coordinates of the center of mass of the thin uniform wire are

⎟⎟⎠

⎞⎜⎜⎝

⎛0 ,sin

m

m

θθR .

For radians 180m πθ =°= :

0sincm ==

ππRx as expected.

Chapter 5

506

For radians 2

90mπθ =°= :

ππ

πR

Rx 2

2

2sin

cm =⎟⎠⎞

⎜⎝⎛

= as expected.

112 ••• A long, thin wire of length L has a linear mass density given by A Bx− , where A and B are positive constants and x is the distance from the more massive end. (a) A condition for this problem to be realistic is that A BL> . Explain why. (b) Determine xcm in terms of L, A, and B. Does your answer makes sense if B = 0? Explain. Picture the Problem The pictorial representation summarizes the information we’re given about the long thin wire. We can use the definition of the center of mass for a continuous object to find the center of mass of the non-uniform rod.

0 x Lλ λ λ( ) A=0

dmdx

( ) BLAL −=( ) BxAx −=

(a) At the end, the density has to be positive, so 0 >− BLA or BLA > . (b) The x coordinate of the center of mass of the non-uniform rod is given by:

∫∫=

dm

xdmxcm (1)

By symmetry:

0cm =y

The density of the long thin wire decreases with distance according to:

( )dxBxAdm −=

Substituting for dm in equation (1) yields:

( )

( )∫

∫

−

−= L

L

dxBxA

dxBxAxx

0

0cm

Evaluate these integrals to obtain:

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−

−=

ABL

ABL

Lx

21

321

2cm


507

The coordinates of the center of mass of the long thin wire are

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−

−0 ,

21

321

2A

BLA

BLL .

Because BLA > , both the numerator and denominator are positive. Because the denominator is always larger than the numerator, it follows that Lx 2

1cm < . This

makes physical sense because the mass of the rod decreases with distance and so most of it is to the left of the midpoint of the rod. Note also that if B = 0, our result predicts a uniform density (of A) and the center of mass is at the midpoint of the rod (as it should be). Motion of the Center of Mass 113 • [SSM] Two 3.0-kg particles have velocities

( ) ( ) jivrrr

m/s 0.3m/s 0.21 += and ( ) ( ) jivrrr

m/s 0.6m/s 0.41 −= . Find the velocity of the center of mass of the system. Picture the Problem We can use Equation 5-24 ( ∑=

iiicm vvrr

mM ) to find the

velocity of the center of mass of this system of two particles. From Equation 5-24:

21

2211iii

cm mmmm

M

m

++

==∑ vv

vv

rrr

r

Substitute numerical values and evaluate cmv

r :

( ) ( ) ( )[ ] ( ) ( ) ( )[ ]

( ) ( ) ji

jijiv

ˆm/s5.1ˆm/s0.3

kg0.3kg0.3

ˆm/s 0.6ˆm/s 0.4kg0.3ˆm/s 0.3ˆm/s 0.2kg0.3cm

−=

+−++

=r

114 • A 1500-kg car is moving westward with a speed of 20.0 m/s, and a 3000-kg truck is traveling east with a speed of 16.0 m/s. Find the velocity of the center of mass of the car-truck system. Picture the Problem Choose a coordinate system in which east is the +x direction and use Equation 5-24 ( ∑=

iiicm vvrr

mM ) to find the velocity of the center of mass

of this system of the two cars.

Chapter 5

508

From Equation 5-24:

ct

ccttiii

cm mmmm

M

m

++

==∑ vv

vv

rrr

r

Express the velocity of the truck: ( )iv ˆm/s0.16t =

r

Express the velocity of the car: ( )iv ˆm/s0.20c −=r

Substitute numerical values and evaluate cmvr :

( )( ) ( )( ) ( ) iiiv ˆm/s00.4

kg1500kg3000

ˆm/s20.0kg1500ˆm/s16.0kg3000cm =

+−+

=r

115 • A force F

r = 12 N i is applied to the 3.0-kg ball in Figure 5-78 in

Problem 103. (No forces act on the other two balls.) What is the acceleration of the center of mass of the three-ball system? Picture the Problem The acceleration of the center of mass of the ball is related to the net external force through Newton’s second law: cmextnet, aF

rrM= .

Use Newton’s second law to express the acceleration of the ball:

Mextnet,

cm

Fa

rr

=

Substitute numerical values and evaluate cma

r : ( )

( )i

ia

ˆm/s4.2

kg1.0kg1.0kg0.3

ˆN12

2

cm

=

++=

r

116 •• A block of mass m is attached to a string and suspended inside an otherwise empty box of mass M. The box rests on a scale that measures the system’s weight. (a) If the string breaks, does the reading on the scale change? Explain your reasoning. (b) Assume that the string breaks and the mass m falls with constant acceleration g. Find the magnitude and direction of the acceleration of the center of mass of the box–block system. (c) Using the result from (b), determine the reading on the scale while m is in free-fall. Picture the Problem Choose a coordinate system in which upward is the +y direction. We can use Newton’s second law, ,cmextnet, aF

rrM= to find the

acceleration of the center of mass of this two-body system.


509

(a) Yes; initially the scale reads (M + m)g. While the block is in free fall, the reading is Mg. (b) Using Newton’s second law, express the acceleration of the center of mass of the system:

tot

extnet,cm m

Fa

rr

=

The box has zero acceleration. Hence:

( ) jja ˆˆ0cm ⎟

⎠⎞

⎜⎝⎛

+−=⎟

⎠⎞

⎜⎝⎛

++

−=mM

mgmM

Mmgr

(c) Let Fn be the normal force exerted by the scale (the scale reading) on the box. Use Newton’s second law to express the net force acting on the system while the block is falling:

cmnextnet, )( amMMgmgFF +=−−=

Solving for Fn yields:

MgmgamMF +++= cmn )(

Substitute for acm and simplify to obtain:

( )

Mg

gmMmM

mgmMF

=

++⎟⎠⎞

⎜⎝⎛

+−+= )(n

as expected, given our answer to (a). 117 •• [SSM] The bottom end of a massless, vertical spring of force constant k rests on a scale and the top end is attached to a massless cup, as in Figure 5-81. Place a ball of mass mb gently into the cup and ease it down into an equilibrium position where it sits at rest in the cup. (a) Draw the separate free-body diagrams for the ball and the spring. (b) Show that in this situation, the spring compression d is given by d = mbg/k. (c) What is the scale reading under these conditions? Picture the Problem (b) We can apply Newton’s second law to the ball to find an expression for the spring’s compression when the ball is at rest in the cup. (c) The scale reading is the force exerted by the spring on the scale and can be found from the application of Newton’s second law to the cup (considered as part of the spring). (a) The free-body diagrams for the ball and spring follow. Note that, because the ball has been eased down into the cup, both its speed and acceleration are zero.

Chapter 5

510

bm k

ballon springby F

r

ballon Earth by F

r

springon ballby F

r

springon

scaleby Fr

spring. theofpart as cup heConsider t

(b) Letting the upward direction be the positive y direction, apply ∑ = yy maF to the ball when it is at

rest in the cup and the spring has been compressed a distance d:

yamFF bballon

Earthby ballon springby =−


ballon Earthby

ballon springby =− FF (1)

Because kdF =ballon springby and

gmF bballon Earthby = :

0b =− gmkd ⇒ k

gmd b=

(c) Apply ∑ = yy maF to the spring:

yamFF spring

springon ballby

springon scaleby =− (2)


springon ballby

springon scaleby =− FF

Because

springon ballby

ballon springby FF = , adding

equations (1) and (2) yields:

0ballon Earthby

springon scaleby =− FF

Solving for springon scaleby F yields: gmFF b

ballon Earthby

springon scaleby ==

118 ••• In the Atwood’s machine in Figure 5-82 the string passes over a fixed cylinder of mass mc. The cylinder does not rotate. Instead, the string slides on its frictionless surface. (a) Find the acceleration of the center of mass of the two-block-cylinder-string system. (b) Use Newton’s second law for systems to find the force F exerted by the support. (c) Find the tension T in the string connecting the blocks and show that F = mcg + 2T.


511

Picture the Problem Assume that the object whose mass is m1 is moving downward and take that direction to be the positive direction. We’ll use Newton’s second law for a system of particles to relate the acceleration of the center of mass to the acceleration of the individual particles. (a) Relate the acceleration of the center of mass to m1, m2, mc and their accelerations:

cc2211cm aaaarrrr mmmM ++=

Because m1 and m2 have a common acceleration of magnitude a and ac = 0:

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−=

c21

21cm mmm

mmaa

From Problem 4-84 we have:: 21

21

mmmm

ga+−

=

Substitute to obtain:

( )( )( ) g

mmmmmmm

mmmmmg

mmmma

c2121

221

c21

21

21

21cm

+++−

=

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=

(b) Use Newton’s second law for a system of particles to obtain:

cmMaMgF =+− where M = m1 + m2 + mc and F is upwards.

Solve for F, substitute for acm from Part (a), and simplify to obtain:

( )

gmmmmm

gmmmmMgMaMgF

⎥⎦

⎤⎢⎣

⎡+

+=

+−

−=−=

c21

21

21

221

cm

4

(c) From Problem 4-84 we have: g

mmmm

T21

212+

=

Substitute in our result from Part (b) and simplify to obtain:

gmTgmgTgm

mmmmF ccc

21

21 2222 +=⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+

+=

Chapter 5

512

119 ••• Starting with the equilibrium situation in Problem 117, the whole system (scale spring, cup, and ball) is now subjected to an upward acceleration of magnitude a (for example, in an elevator). Repeat the free-body diagrams and calculations in Problem 117. Picture the Problem Because the whole system is accelerating upward, the net upward force acting on the system must be upward. Because the spring is massless, the two forces acting on it remain equal and are oppositely directed. (b) We can apply Newton’s second law to the ball to find an expression for the spring’s compression under the given conditions. (c) The scale reading, as in Problem 117, is the force the scale exerts on the spring and can be found from the application of Newton’s second law to the spring. (a) The free-body diagrams for the ball and spring follow. Note that, because the system is accelerating upward,

ballon Earthby

ballon springby FF > whereas

springon ballby

springon scaleby FF = .

bm kspring. theofpart

as cup heConsider t

springon

scaleby Fr

springon

ballby Fr

ballon springby F

r

ballon Earth by F

r

(b) Letting the upward direction be the positive y direction, apply ∑ = yy maF to the ball when the spring is compressed a distance d' :

yamFF bballon Earthby

ballon springby =− (1)

Because 'ballon springby kdF = , ay = a, and

gmF bballon Earthby = :

amgmkd' bb =− ⇒( )k

agmd' += b

(c) Apply ∑ = yy maF to the spring: 0springon ballby

springon scaleby =− FF (2)


513

Because

springon ballby

ballon springby FF = and

ay = a, adding equations (1) and (2) yields:

amFF bballon Earthby

springon scaleby =−

Solving forspringon scaleby F yields:

ballon Earthby b

springon scaleby FamF +=

Substitute forballon Earthby F and simplify to

obtain:

( )gamF += bspringon scaleby

Remarks: Note that the two forces acting on the spring and the upward force acting on the ball, while still equal, are larger (because the system is accelerating upward) than they were in Problem 117. General Problems 120 • In designing your new house in California, you are prepared for it to withstand a maximum horizontal acceleration of 0.50g. What is the minimum coefficient of static friction between the floor and your prized Tuscan vase so that the vase does not slip on the floor under these conditions? Picture the Problem The forces acting on the vase are the gravitational force

gFrr

m=g exerted by Earth, the normal

force nFr

exerted by the floor, and the static friction force sf

r, also exerted by

the floor. We can apply Newton’s second law to find the minimum coefficient of static friction that will prevent the vase from slipping.

x

y

nFr

gmFrr

=g

sfr

Apply aF

rrm=∑ to the vase to

obtain:

xx mafF ==∑ s (1) and

0n =−=∑ mgFFy

Relate the static friction force sfr

to the minimum coefficient of static friction μs,min that will prevent the vase from slipping:

nmins,s Ff μ= or, because Fn = mg,

mgf mins,s μ=

Chapter 5

514

Substituting for fs in equation (1) yields:

xmamg =mins,μ ⇒gax=mins,μ

Because gax 50.0= : 50.050.0mins, ==

ggμ

121 • A 4.5-kg block slides down an inclined plane that makes an angle of 28º with the horizontal. Starting from rest, the block slides a distance of 2.4 m in 5.2 s. Find the coefficient of kinetic friction between the block and plane. Picture the Problem The forces that act on the block as it slides down the incline are shown on the free-body diagram to the right. The acceleration of the block can be determined from the distance-and-time information given in the problem. The application of Newton’s second law to the block will lead to an expression for the coefficient of kinetic friction as a function of the block’s acceleration and the angle of the incline.

nFr

θx

y

kfr

gmFrr

=g

m

Apply ∑ = aF

rrm to the block: xx mafmgF∑ =−= ksinθ (1)


Set fk = μkFn in equation (1) to obtain: xmaFmg =− nksin μθ (3)

Solve equation (2) for Fn and substitute in equation (3) to obtain:

xmamgmg =− θμθ cossin k

Solving for μk yields: θ

θμcos

sink g

ag x−= (4)

Using a constant-acceleration equation, relate the distance the block slides to its sliding time:

( )221

0 ΔΔΔ tatvx xx += or, because v0x = 0,

( )221 ΔΔ tax x= ⇒

( )2ΔΔ2txax =


515

Substitute for ax in equation (4) to obtain:

( )θ

θμ

cosΔΔ2sin 2

k gtxg −

=

Substitute numerical values and evaluate μk:

( ) ( )( )

( ) 51.028cosm/s 81.9

s 2.5m 4.2228sinm/s 81.9

2

22

k =°

−°=μ

122 •• You plan to fly a model airplane of mass 0.400 kg that is attached to a horizontal string. The plane will travel in a horizontal circle of radius 5.70 m. (Assume the weight of the plane is balanced by the upward ″lift″ force of the air on the wings of the plane.) The plane will make 1.20 revolutions every 4.00 s. (a) Find the speed at which you must fly the plane. (b) Find the force exerted on your hand as you hold the string (assume the string is massless). Picture the Problem The force exerted on your hand as you hold the string is the reaction force to the tension F

r in

the string and, hence, has the same magnitude. The speed of the plane can be calculated from the data concerning the radius of its path and the time it takes to make one revolution. The application of Newton’s second law will give us the tension F in the string.

x

y

gmFrr

=g

liftFr

Fr

(a) Express the speed of the airplane in terms of the circumference of the circle in which it is flying and its period:

Trv π2

=


( ) m/s10.7

rev1.20s4.00m5.702π

==v

(b) Apply∑ = xx maF to the model

airplane:

rvmF

2

=

Chapter 5

516


( )( ) N8.10m 70.5

m/s 0.741kg0.400 2

==F

123 •• Your moving company is to load a crate of books on a truck with the help of some planks that slope upward at 30.0º. The mass of the crate is 100 kg, and the coefficient of sliding friction between it and the planks is 0.500. You and your employees push horizontally with a combined net force F

r. Once the crate

has started to move, how large must F be in order to keep the crate moving at constant speed? Picture the Problem The free-body diagram shows the forces acting on the crate of books. The kinetic friction force opposes the motion of the crate up the incline. Because the crate is moving at constant speed in a straight line, its acceleration is zero. We can determine F by applying Newton’s second law to the crate, substituting for fk, eliminating the normal force, and solving for the required force.

gFr

nFr

θ

x

y

kfr F

rθ

Apply∑ = aF

rrm to the crate, with

both ax and ay equal to zero, to the crate:

∑ =−−= 0sincos k θθ mgfFFx

and ∑ =−−= 0cossinn θθ mgFFFy

Because ,nkk Ff μ= the x-equation becomes:

0sincos nk =−− θμθ mgFF (1)

Solving the y-equation for Fn yields:

θθ cossinn mgFF +=

Substitute for Fn in equation (1) and solve for F to obtain:

( )θμθθμθ

sincoscossin

k

k

−+

=mgF


( )( ) ( )( )( ) kN49.1

sin300.500cos30.0cos30.00.500sin30m/s9.81kg100 2

=°−°

°+°=F


517

124 •• Three forces act on an object in static equilibrium (Figure 5-83). (a) If F1, F2, and F3 represent the magnitudes of the forces acting on the object, show that F1/sin θ23 = F2/sin θ31 = F3/sin θ12. (b) Show that 2 2 2

1 2 3 2 3 232 cosF F F F F θ= + + . Picture the Problem The fact that the object is in static equilibrium under the influence of the three forces means that .0321 =++ FFF

rrr Drawing the

corresponding force triangle will allow us to relate the forces to the angles between them through the law of sines and the law of cosines. (a) Using the fact that the object is in static equilibrium, redraw the force diagram connecting the forces head-to-tail:

Appling the law of sines to the triangle yields: ( ) ( ) ( )12

3

13

2

23

1

sinsinsin θπθπθπ −=

−=

−FFF

Use the trigonometric identity sin(π − α) = sinα to obtain:

12

3

13

2

23

1

sinsinsin θθθFFF

==

(b) Appling the law of cosines to the triangle yields:

( )23322

32

22

1 cos2 θπ −−+= FFFFF

Use the trigonometric identity cos(π − α) = −cosα to obtain: 2332

23

22

21 cos2 θFFFFF ++=

125 •• In a carnival ride, you sit on a seat in a compartment that rotates with constant speed in a vertical circle of radius 5.0 m. The ride is designed so your head always points toward the center of the circle. (a) If the ride completes one full circle in 2.0 s, find the direction and magnitude of your acceleration. (b) Find the slowest rate of rotation (in other words, the longest time Tm to complete one full circle) if the seat belt is to exert no force on you at the top of the ride. Picture the Problem We can calculate your acceleration from your speed that, in turn, is a function of the period of the motion. To determine the longest period of the motion, we focus our attention on the situation at the very top of the ride when the seat belt is exerting no force on you. We can use Newton’s second law to relate the period of the motion to your acceleration and speed.

Chapter 5

518

r

gmr

(a) Because the motion is at constant speed, your acceleration is entirely radial and is given by:

rva

2

c =

Express your speed as a function of the radius of the circle and the period of the motion:

Trv π2

=

Substitute for v in the expression for ac and simplify to obtain: 2

2

c4T

ra π=


( )( )

22

2

c m/s49s0.2

m0.54==

πa

(b) Apply ∑ = aF

rrm to yourself

when you are at the top of the circular path and the seat belt is exerting no force on you:

∑ ==rvmmgF

2

r ⇒ grv =

Express the period of your motion as a function of the radius of the circle:

vrT π2

m =

Substituting for v and simplifying yields:

gr

grrT ππ 22

m ==

The slowest rate of rotation is the reciprocal of Tm:

rgT

π211

m =−


519

Substitute numerical values and evaluate 1

m−T :

rev/min 13min

s 60s 2229.0

m5.0m/s9.81

21

1

21

m

≈

×=

=

−

−

πT

Remarks: You are ″weightless″ under the conditions described in Part (b). 126 •• A flat-topped toy cart moves on frictionless wheels, pulled by a rope under tension T. The mass of the cart is m1. A load of mass m2 rests on top of the cart with the coefficient of static friction μs between the cart and the load. The cart is pulled up a ramp that is inclined at angle θ above the horizontal. The rope is parallel to the ramp. What is the maximum tension T that can be applied without causing the load to slip? Picture the Problem The pictorial representation to the right shows the cart and its load on the inclined plane. The load will not slip provided its maximum acceleration is not exceeded. We can find that maximum acceleration by applying Newton’s second law to the load. We can then apply Newton’s second law to the cart-plus-load system to determine the tension in the rope when the system is experiencing its maximum acceleration.

Tr

θ

1m2m

Draw the free-body diagram for the cart and its load:

x

y

Tr

( )gmmr

21 +θ

1m

n,1Fr

Apply ∑ = xx maF to the cart plus

its load:

( )( ) max,21

21 sin

xammgmmT+=

+− θ (1)

Chapter 5

520

Draw the free-body diagram for the load of mass m2 on top of the cart:

x

y

n,2Fr

θ

sfr

gmr

2

2m

Apply ∑ = aFrr

m to the load on top

of the cart: ∑ =−= max,22maxs, sin xx amgmfF θ

and 0cos22,n =−=∑ θgmFFy

Because ,n,2smaxs, Ff μ= the x

equation becomes:

max,22n,2s sin xamgmF =− θμ (2)

Solving the y equation for Fn,2 yields:

θcos2n,2 gmF =

Substitute for Fn,2 in equation (2) to obtain:

max,222s sincos xamgmgm =− θθμ

Solving for max,xa and simplifying

yields:

( )θθμ sincossmax, −= gax (3)

Substitute for max,xa in equation (1)

and solve for T to obtain: ( ) θμ coss21 gmmT +=

127 ••• A sled weighing 200 N that is held in place by static friction, rests on a 15º incline (Figure 5-84). The coefficient of static friction between the sled and the incline is 0.50. (a) What is the magnitude of the normal force on the sled? (b) What is the magnitude of the static frictional force on the sled? (c) The sled is now pulled up the incline at constant speed by a child walking up the incline ahead of the sled. The child weighs 500 N and pulls on the rope with a constant force of 100 N. The rope makes an angle of 30º with the incline and has negligible mass. What is the magnitude of the kinetic frictional force on the sled? (d) What is the coefficient of kinetic friction between the sled and the incline? (e) What is the magnitude of the force exerted on the child by the incline?


521

Picture the Problem The free-body diagram for the sled while it is held stationary by the static friction force is shown to the right. We can solve this problem by repeatedly applying Newton’s second law under the conditions specified in each part of the problem.

n,1Fr

gmr

1

sfr

θ

x

y

(a) Apply∑ = yy maF to the sled: 0cos1n,1 =− θgmF

Solve for Fn,1:

θcos1n,1 gmF =

Substitute numerical values and evaluate Fn,1:

( ) kN19.015cosN200n,1 =°=F

(b) Apply∑ = xx maF to the sled: 0sin1s =− θgmf ⇒ θsin1s gmf =


( ) N5215sinN200s =°=f

(c) Draw the free-body diagram for the sled when it is moving up the incline at constant speed: n,1F

r

gmr

1 θ

x

y

kfr

Fr

°30

Apply 0=∑ xF to the sled to obtain: 0sin30cos k1 =−−° fgmF θ Solving for kf yields:

θsin30cos 1k gmFf −°=

Substitute numerical values and evaluate kf :

( ) ( )N 35N 84.34

15sinN20030cosN100

==

°−°=kf

Chapter 5

522

(d) The coefficient of kinetic friction between the sled and the incline is given by:

n,1Ffk

k =μ (1)

Apply 0=∑ yF to the sled to obtain:

0cos30sin 1n,1 =−°+ θgmFF

Solve for n,1F :

°−= 30sincos1n,1 FgmF θ

Substituting for n,1F in equation (1)

yields: °−=

30sincos1 Fgmfk

k θμ

Substitute numerical values and evaluate kμ :

( ) ( )

24.0

30sinN 10015cosN 200N 84.34

=

°−°=kμ

(e) Draw the free body diagram for the child:

θ

n,2Fr

gmr

2

Fr

−maxs,fr

°30

x

y

Express the net force Fc exerted on the child by the incline:

2maxs,

2n2c fFF += (1)

Noting that the child is stationary, apply∑ = aF

rrm to the child:

0

15sin30cos 2maxs,

=

°−°−=∑ gmFfFx

and 030sin15cos2n2 =°−°−=∑ FgmFFy

Solve the x equation for fs,max and the y equation for Fn,2:

°+°= 15sin30cos 2maxs, gmFf

and °+°= 30sin15cos2n,2 FgmF


523

Substitute numerical values and evaluate fs,max and Fn,2:

( ) ( )N0.216

15sinN50030cosN100maxs,

=

°+°=f

and ( ) ( )

N0.53330sinN10015cosN500n,2

=

°+°=F

Substitute numerical values in equation (1) and evaluate F:

( ) ( )kN58.0

N0.533N0.216 22c

=

+=F

128 •• In 1976, Gerard O’Neill proposed that large space stations be built for human habitation in orbit around Earth and the moon. Because prolonged free-fall has adverse medical effects, he proposed making the stations in the form of long cylinders and spinning them around the cylinder axis to provide the inhabitants with the sensation of gravity. One such O’Neill colony is to be built 5.0 miles long, with a diameter of 0.60 mi. A worker on the inside of the colony would experience a sense of ″gravity,″ because he would be in an accelerated frame of reference due to the rotation. (a) Show that the ″acceleration of gravity″ experienced by the worker in the O’Neill colony is equal to his centripetal acceleration. (Hint: Consider someone ″looking in″ from outside the colony.) (b) If we assume that the space station is composed of several decks that are at varying distances (radii) from the axis of rotation, show that the ″acceleration of gravity″ becomes weaker the closer the worker gets to the axis. (c) How many revolutions per minute would this space station have to make to give an ″acceleration of gravity″ of 9.8 m/s2 at the outermost edge of the station? Picture the Problem Let v represent the speed of rotation of the station, and r the distance from the center of the station. Because the O’Neill colony is, presumably, in deep space, the only acceleration one would experience in it would be that due to its rotation. (a) The acceleration of anyone who is standing inside the station is a = v2/r . This acceleration is directed toward the axis of rotation. If someone inside the station drops an apple, the apple will not have any forces acting on it once released, but will move along a straight line at constant speed. However, from the point of view of our observer inside the station, if he views himself as unmoving, the apple is perceived to have an acceleration of v2/r directed away from the axis of rotation (a "centrifugal" acceleration). (b) Each deck must rotate about the central axis with the same period T. Relate the speed of a person on a particular deck to his/her distance r from the center:

Trv π2

=

Chapter 5

524

Express the "acceleration of gravity" perceived by someone a distance r from the center:

2

22

c4T

rr

va π== , a result that tells us

that the ″acceleration of gravity″ decreases as r decreases.

(c) Relate the desired acceleration to the radius of the space station and its period:

2

24T

ra π= ⇒

arT

24π=


min 0.74 s44m/s8.9

mikm1.609mi30.04

2

2

≈=

⎟⎠⎞

⎜⎝⎛ ×

=π

T

Take the reciprocal of this time to find the number of revolutions per minute Babylon 5 has to make in order to provide this ″Earth-like″ acceleration:

min/rev4.11 =−T

129 •• A child of mass m slides down a slide inclined at 30º in time t1. The coefficient of kinetic friction between her and the slide is μk. She finds that if she sits on a small sled (also of mass m) with frictionless runners, she slides down the same slide in time 12

1 t . Find μk. Picture the Problem The following free-body diagram shows the forces acting on the child as she slides down the incline. We’ll first use Newton’s second law to derive an expression for μk in terms of her acceleration and then use Newton’s second law to find her acceleration when riding the frictionless cart. Using a constant-acceleration equation, we’ll relate these two accelerations to her descent times and solve for her acceleration when sliding. Finally, we can use this acceleration in the expression for μk.

nFr

θx

y

kfr

gmFrr

=g


525

Apply ∑ = aFrr

m to the child as she

slides down the incline: ∑ =−= xx mafmgF ,1ksinθ

and ∑ =−= 0cosn θmgFFy

Because nkk Ff μ= , the x-equation can be written:

xmaFmg ,1nksin =− μθ (1)

Solving the y-equation for nF yields:

θcosn mgF =

Substitute for nF in equation (1) to obtain:

xmamgmg ,1k cossin =− θμθ

Solving for μk yields: °

−°=30cos

30tan ,1k g

a xμ (2)

Apply∑ = xx maF to the child as

she rides the frictionless cart down the incline and solve for her acceleration xa ,2 :

xmamg ,230sin =°

and °= 30sin,2 ga x

Letting s represent the distance she slides down the incline, use a constant-acceleration equation to relate her sliding times to her accelerations and distance traveled down the slide :

0where 021,12

110 =+= xxx vtatvs

and 0where 0

22,22

120 =+= xxx vtatvs

Equate these expressions, substitute t2 = 2

1 t1 and solve for a1,x:

°== 30sin41

,241

,1 gaa xx

Substitute for a1,x in equation (2) to obtain:

°=°°

−°= 30tan30cos30sin

30tan 434

1

k gg

μ

Substitute numerical values and evaluate μk:

43.030tan43

k =°=μ

130 ••• The position of a particle of mass m = 0.80 kg as a function of time is given by ( ) ( ) jijir ˆcosˆsinˆˆ tRtRyx ωω +=+=

r , where R = 4.0 m and ω = 2π s–1. (a) Show that the path of this particle is a circle of radius R, with its center at the origin of the xy plane. (b) Compute the velocity vector. Show that vx/vy = –y/x.

Chapter 5

526

(c) Compute the acceleration vector and show that it is directed toward the origin and has the magnitude v2/R. (d) Find the magnitude and direction of the net force acting on the particle. Picture the Problem The path of the particle is a circle if r is constant. Once we have shown that it is, we can calculate its value from its components and determine the particle’s velocity and acceleration by differentiation. The direction of the net force acting on the particle can be determined from the direction of its acceleration. (a) Express the magnitude of r

r in

terms of its components:

22yx rrr +=

Evaluate r with tRrx ωsin= and tRry ωcos= :

[ ] [ ]( ) RttR

tRtRr

=+=

+=

ωω

ωω222

22

cossin

cossin

This result shows that the path of the particle is a circle of radius R centered at the origin. (b) Differentiate r

rwith respect to time to obtain :v

r

[ ] [ ]

( )[ ] ( )[ ] ji

jirv

ˆm/s2sin0.8ˆm/s2cos0.8

ˆsinˆcos

tt

tRtRdtd

ππππ

ωωωω

−=

−+==r

r

Express the ratio :y

x

vv t

tt

vv

y

x ωωπωπ cot

sin0.8cos0.8

−=−

= (1)

Express the ratio :xy

− ttRtR

xy ω

ωω cot

sincos

−=−=− (2)

From equations (1) and (2) we have:

xy

vv

y

x −=

(c) Differentiate v

r with respect to time to obtain :ar

( )[ ] ( )[ ]jiva ˆcosm/s16ˆsinm/s16 2222 ttdtd ωπωπ −+−==r

r


527

Factor −4.0π2/s2 from ar to obtain:

( ) ( ) ( )[ ] ( )rjtitarr 2222 s 0.4ˆcos0.4ˆsin0.4s 0.4 −− −=+−= πωωπ

Because a

r is in the opposite direction from ,rr it is directed toward the center of

the circle in which the particle is traveling.

Find the ratio Rv2

: ( ) 2222

m/s16m0.4

m/s0.8 ππ===

Rva

(d) Apply∑ = aF

rrm to the particle:

( )( )N13

m/s16kg80.02

22net

π

π

=

== maF

Because the direction of netF

ris the same as that of a

r , netFr

is toward the center of the circle. 131 ••• You are on an amusement park ride with your back against the wall of a spinning vertical cylinder. The floor falls away and you are held up by static friction. Assume your mass is 75 kg. (a) Draw a free-body diagram of yourself. (b) Use this diagram with Newton’s laws to determine the force of friction on you. (c) The radius of the cylinder is 4.0 m and the coefficient of static friction between you and the wall is 0.55. What is the minimum number of revolutions per minute necessary to prevent you from sliding down the wall? Does this answer hold only for you? Will other, more massive, patrons fall downward? Explain. Picture the Problem The application of Newton’s second law and the definition of the maximum static friction force will be used to determine the period T of the motion. The reciprocal of the period will give us the minimum number of revolutions required per unit time to hold you in place. (a) The free-body diagram showing the forces acting on you when you are being held in place by the maximum static friction force is shown to the right.

nFr

Fr

g

maxs,fr

y

xm

Chapter 5

528

(b) Apply ∑ = aF

rrm to yourself

while you are held in place by friction:

∑ ==r

vmFFx

2

n (1)

and ∑ =−= 0maxs, mgfFy (2)

Solve equation (2) for maxs,f : mgf =maxs,

Substitute numerical values and evaluate maxs,f :

( )( ) kN 74.0m/s 81.9kg 75 2maxs, ==f

(c) The number of revolutions per minute N is the reciprocal of the period in minutes:

TN 1= (3)

Because nsmaxs, Ff μ= , equation (1)

can be written:

rvmmgf 2

ss

maxs, ==μμ

(4)

Your speed is related to the period of your motion: T

rv π2=

Substitute for v in equation (4) to obtain: 2

2

2

s

42

Tmr

rT

r

mmg ππ

μ=

⎟⎠⎞

⎜⎝⎛

=

Solving for T yields: g

rT s2 μπ=

Substitute for T in equation (3) to obtain:

rg

gr

Nss 2

1

2

1μπμπ

==

Substitute numerical values and evaluate N: ( )( )

rev/min 20

rev/s 336.0m 0.455.0

m/s 81.921 2

=

==π

N

Because your mass does not appear in the expression for N, this result holds for all patrons, regardless of their mass.


529

132 ••• A block of mass m1 is on a horizontal table. The block is attached to a 2.5-kg block (m2) by a light string that passes over a pulley at the edge of the table. The block of mass m2 dangles 1.5 m above the ground (Figure 5-85). The string that connects them passes over a frictionless, massless pulley. This system is released from rest at t = 0 and the 2.5-kg block strikes the ground at t = 0.82 s. The system is now placed in its initial configuration and a 1.2-kg block is placed on top of the block of mass m1. Released from rest, the 2.5-kg block now strikes the ground 1.3 s later. Determine the mass m1 and the coefficient of kinetic friction between the block whose mass is m1 and the table. Picture the Problem The free-body diagrams below show the forces acting on the blocks whose masses are m1 and m2. The application of Newton’s second law and the use of a constant-acceleration equation will allow us to find a relationship between the coefficient of kinetic friction and m1. The repetition of this procedure with the additional block on top of the block whose mass is m1 will lead us to a second equation that, when solved simultaneously with the former equation, leads to a quadratic equation in m1. Finally, its solution will allow us to substitute in an expression for μk and determine its value.

y

n,1Fr

1Tr

gmr

1

kfr

1m

2Tr

gmr

2

2m

x

x

Using a constant-acceleration equation, relate the displacement of the system in its first configuration as a function of its acceleration and fall time:

( )2121

0 ΔΔΔ tatvx xx += or, because v0x = 0,

( )21,21 ΔΔ tax x= ⇒

( )21, ΔΔ2txax =

Substitute numerical values and evaluate 1,xa :

( )( )

221, m/s4616.4

s82.0m5.12

==xa


whose mass is m2:

1,212 xamTgm =− ⇒ ( )xagmT 121 −=

Chapter 5

530

Substitute numerical values and evaluate T1:

( )( )N371.13

m/s4616.4m/s81.9kg5.2 221

=−=T

Apply ∑ = aF

rrm to the block whose

mass is m1: ∑ =−= 1,1k1 xx amfTF

and ∑ =−= 01n,1 gmFFy

Using fk = μkFn, eliminate Fn between the two equations to obtain:

1,11k1 xamgmT =− μ (1)

Find the acceleration 2,xa for the

second run: ( )( )( )

2222, m/s775.1

s3.1m5.12

ΔΔ2

===txax

T2 is given by: ( )

( )( )N1.20

m/s775.1m/s81.9kg5.2 22

2,22

=−=

−= xagmT

Apply ∑ = xx maF to the system

with the 1.2-kg block in place:

( )( ) 2,1

1k2

kg2.1kg2.1

xamgmT

+=+− μ

(2)

Solve equation (1) for μk: gmamT x

1

1,11k

−=μ (3)

Substitute for μk in equation (2) and simplify to obtain the quadratic equation in m1:

005.16940.9686.2 121 =−+ mm

Use your graphing calculator or the quadratic formula to obtain:

kg2.1kg215.11 ==m

Substitute numerical values in equation (3) and evaluate μk:

( )( )( )( ) 67.0

m/s81.9kg1.215m/s1646.4kg 1.215N375.13

2

2

k =−

=μ

133 ••• Sally claims flying squirrels do not really fly; they jump and use folds of skin that connect their forelegs and their back legs like a parachute to allow them to glide from tree to tree. Liz decides to test Sally’s hypothesis by calculating the terminal speed of a falling outstretched flying squirrel. If the constant b in the drag force is proportional to the area of the object facing the air


531

flow, use the results of Example 5-8 and some assumptions about the size of the squirrel to estimate its terminal (downward) speed. Is Sally’s claim supported by Liz’s calculation? Picture the Problem A free-body diagram showing the forces acting on the squirrel under terminal-speed conditions is shown to the right. We’ll assume that b is proportional to the squirrel’s frontal area and that this area is about 0.1 that of a human being. Further, we’ll assume that the mass of a squirrel is about 1.0 kg. Applying Newton’s second law will lead us to an expression for the squirrel’s terminal speed.

jbvF ˆ2tdrag =

r

y

jmgF ˆg −=r

Apply yy maF =∑ to the squirrel:

ymaFF =− gd or, because ay = 0,

0gd =− FF

Substituting for Fd and Fg yields: 02

t =−mgbv ⇒b

mgv =t (1)

Assuming that Asquirrel = (0.1)Ahuman:

( ) human1.0 bb =

From Example 5-8:

kg/m 251.0human =b

Substitute numerical values in equation (1) and evaluate vt:

( )( )( )( ) m/s 20

kg/m 251.0)1.0m/s 81.9kg 0.1 2

t ≈=v

Because the squirrel’s terminal speed is approximately 80 km/h and this value is less than half that of the skydiver in Example 5-8, Sally’s claim seems to be supported by Liz’s calculation. 134 ••• After a parachutist jumps from an airplane (but before he pulls the rip cord to open his parachute), a downward speed of up to 180 km/h can be reached. When the parachute is finally opened, the drag force is increased by about a factor of 10, and this can create a large jolt on the jumper. Suppose this jumper falls at 180 km/h before opening his chute. (a) Determine the parachutist’s acceleration when the chute is just opened, assuming his mass is 60 kg. (b) If rapid accelerations greater than 5.0g can harm the structure of the human body, is this a safe practice?

Chapter 5

532

Picture the Problem The free-body diagram shows the drag force dF

r

exerted by the air and the gravitational force gF

rexerted by

Earth acting on on the parachutist just after his chute has opened. We can apply Newton’s second law to the parachutist to obtain an expression for his acceleration as a function of his speed and then evaluate this expression for tvv = .

y

jbvF ˆ10 2d =r

jmgF ˆg −=r

(a) Apply yy maF =∑ to the parachutist immediately after the chute opens:

open chute

210 mamgbv =−

Solving foropen

chutea yields:

gv

mba −= 2

open chute 10 (1)

Before the chute opened:

ymamgbv =−2

Under terminal speed conditions, ay = 0 and:

02t =−mgbv ⇒ 2

tvg

mb=

Substitute for b/m in equation (1) to obtain:

gvv

gvv

ga

⎥⎦

⎤⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−⎟⎟⎠

⎞⎜⎜⎝

⎛=

1110

110

22t

22topen

chute

Evaluating

open chutea for v = vt yields:

ggvv

a 91110 2t2

topen chute =

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛=

(b) Because this acceleration exceeds the safe acceleration by 4g, this is not a safe practice. 135 • Find the location of the center of mass of the Earth–moon system relative to the center of Earth. Is it inside or outside the surface of Earth? Picture the Problem We can use the definition of the location of the center of mass of a system of particles to find the location of the center of mass of the Earth-moon system. The following pictorial representation is, of course, not shown to scale.


533

0

Emm

cmx xx

Earthmoon

m

m

The center of mass of the Earth-moon system is given by: mE

mcm,mEcm,Ecm mm

xmxmx

++

=


( )( ) ( )( ) m 1067.4kg 1036.7kg 1098.5

m 1084.3kg 1036.70kg 1098.5 62224

82224

cm ×=×+×

××+×=x

Because the radius of Earth is 6.37 × 106 m, the center of mass of the Earth-moon system is inside Earth. 136 •• A circular plate of radius R has a circular hole of radius R/2 cut out of it (Figure 5-86). Find the center of mass of the plate after the hole has been cut. Hint: The plate can be modeled as two disks superimposed, with the hole modeled as a disk of negative mass. Picture the Problem By symmetry, xcm = 0. Let σ be the mass per unit area of the disk. The mass of the modified disk is the difference between the mass of the whole disk and the mass that has been removed. Start with the definition of ycm:

hole

holeholediskdisk

hole

iii

cm

mMymym

mM

ymy

−−

=

−=∑

Express the mass of the complete disk:

2rAM σπσ ==

Express the mass of the material removed: Mrrm 4

1241

2

hole 2==⎟

⎠⎞

⎜⎝⎛= σπσπ

Substitute and simplify to obtain: ( ) ( )( )

rMM

rMMy 6

1

41

21

41

cm0

=−

−−=

Chapter 5

534

137 •• An unbalanced baton consists of a 50-cm-long uniform rod of mass 200 g. At one end there is a 10-cm-diameter uniform solid sphere of mass 500 g, and at the other end there is a 8.0-cm-diameter uniform solid sphere of mass 750 g. (The center-to-center distance between the spheres is 59 cm.) (a) Where, relative to the center of the light sphere, is the center of mass of this baton? (b) If this baton is tossed straight up (but spinning) so that its initial center of mass speed is 10.0 m/s, what is the velocity of the center of mass 1.5 s later? (c) What is the net external force on the baton while in the air? (d) What is the baton’s acceleration 1.5 s following its release? Picture the Problem The pictorial representation summarizes the information concerning the unbalanced baton and shows a convenient choice for a coordinate system.

cm ,x

cm ,y

0 555 5930

g 5001 =m g 7503 =mcm 0.51 =r cm 0.43 =r

g 2002 =m

(a) The x coordinate of the center of mass of the unbalanced baton is given by:

321

cm,33cm,22cm,11cm mmm

xmxmxmx

++++

=


( )( ) ( )( ) ( )( ) cm 35g 750g 200g 500

cm 95g 750cm 30g 2000g 500cm =

++++

=x

(b) Because the center of mass of the baton acts like a point particle, use a constant-acceleration equation to express is velocity as a function of time:

( ) tavtv yy += 0, or, because ay = −g, ( ) tgvtv y −= 0,

Substitute numerical values and evaluate v(1.5 s):

( ) ( )( )

downward m/s 7.4

m/s 7.4s 5.1m/s 81.9m/s 0.10s 5.1 2

=

−=−=v

(c) The net external force acting on the baton is the gravitational force (weight) exerted on it by Earth:

( )gmmmFF 321gnet ++==


535

Substitute numerical values and evaluate Fnet:

( )( )kN 2.14

m/s 81.9g 750g 200g 500 2net

=

++=F

(d) Again, because the center of mass acts like a point particle, its acceleration is that of a point particle in flight near the surface of Earth:

2cm m/s 81.9−== ga

138 •• You are standing at the very rear of a 6.0-m-long, 120-kg raft that is at rest in a lake with its prow only 0.50 m from the end of the pier (Figure 5-87). Your mass is 60 kg. Neglect frictional forces between the raft and the water. (a) How far from the end of the pier is the center of mass of the you-raft system? (b) You walk to the front of the raft and then stop. How far from the end of the pier is the center of mass now? (c) When you are at the front of the raft, how far are you from the end of the pier? Picture the Problem Let the origin be at the edge of the pier and the +x direction be to the right as shown in the pictorial representation immediately below.

pier

m ,x0

ym

rm

0.53.56.5

In the following pictorial representation, d is the distance of the end of the raft from the pier after you have walked to its front. The raft moves to the left as you move to the right; with the center of mass of the you-raft system remaining fixed (because Fext,net = 0). The diagram shows the initial (xy,i) and final (xy,f) positions of yourself as well as the initial (xr_cm,i) and final (xr_cm,f) positions of the center of mass of the raft both before and after you have walked to the front of the raft.

pier

m ,x0

ym

rm

dd + 3.0 md + 6.0 m (a) xcm before you walk to the front of the raft is given by:

ry

i r_cm,ry,iycm mm

xmxmx

+

+=

Chapter 5

536


( )( ) ( )( )

m 5.4

kg 120kg 60m 5.3kg 120m 5.6kg 60

cm

=

++

=x

(b) Because Fext,net = 0, the center of mass remains fixed:

m 4.5' cmcm == xx

(c) Call the distance of the raft from the pier after you have walked to the front of the raft d and express the new location cm'x of the center of mass of the system:

( )m 4.5

m 0.3

'

ry

ry

ry

fr_cm,rfy,ycm

=+++

=

++

=

mmdmdm

mmxmxm

x

Solving for d yields: ( ) ( )ry

ry m 5.1m 4.5mm

mmd

++

=


( )( ) ( )( )

m 5.2

kg 120kg 60kg 120m 5.1kg 60m 5.4

=

++

=d

139 •• An Atwood’s machine that has a frictionless massless pulley and massless strings has a 2.00-kg object hanging from one side and 4.00-kg object hanging from the other side. (a) What is the speed of each object 1.50 s after they are released from rest? (b) At that time, what is the velocity of the center of mass of the two objects? (c) At that time, what is the acceleration of the center of mass of the two objects? Picture the Problem The forces acting on the 2.00-kg (m1) and 4.00-kg (m2) objects are shown in the free-body diagrams to the right. Note that the two objects have the same acceleration and common speeds. We can use constant- acceleration equations and Newton’s second law in the analysis of the motion of these objects.

y

1m

1Tr

y

2Tr

2m

gmFrr

1g,1 = gmFrr

2g,2 =


537

(a) Use a constant-acceleration equation to relate the speed of the objects to their common acceleration:

( ) tavtv yy += 0

or, because v0y = 0, ( ) tatv y= (1)

Apply yy maF =∑ to the object

whose mass is m1:

yamgmT 111 =−

Apply yy maF =∑ to the object

whose mass is m2:

yamTgm 222 =−

Because T1 = T2, adding these equations yields:

yy amamgmgm 1212 +=−

Solve for ay to obtain: g

mmmmay ⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

=12

12 (2)

Substituting for ay in equation (1) yields:

( ) tgmmmmtv ⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

=12

12

Substitute numerical values and evaluate ( )s 50.1v :

( ) ( )( ) m/s 91.4m/s 905.4s 50.1m/s 81.9kg 00.2kg 00.4kg 00.2kg 00.4s 50.1 2 ==⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

=v

(b) The velocity of the center of mass is given by: 21

2211cm mm

mm++

=vv

vrr

r

Substitute numerical values and evaluate ( )s 50.1cmv :

( ) ( )( ) ( )( ) ( ) jjjv ˆm/s 64.1kg 00.4kg 00.2

ˆm/s 905.4kg 00.4ˆm/s 905.4kg 00.2s 50.1cm −=+

−+=

The velocity of the center of mass is downward. m/s 64.1

(c) The acceleration of the center of mass is given by:

21

2211cm mm

mm++

=aa

arr

r

Chapter 5

538

Substituting for ay from equation (2) and simplifying yields:

jjj

a ˆˆˆ

2

12

12

21

12

122

12

121

cm gmmmm

mm

gmmmmmg

mmmmm

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−=+

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−+⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=r

Substitute numerical values and evaluate acm:

( ) ( )jja ˆm/s 09.1ˆm/s 81.9kg 00.2kg 00.4kg 00.2kg 00.4 22

2

cm −=⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−=r

The acceleration of the center of mass is downward. m/s 09.1 2

Chapter 5 Additional Applications of Newton’s...

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